It is often said that mathematics is the language of science. If this is true, then the language of mathematics is numbers. The earliest use of numbers occurred 100 centuries ago in the Middle East to count, or enumerate items. Farmers, cattlemen, and tradesmen used tokens, stones, or markers to signify a single quantity—a sheaf of grain, a head of livestock, or a fixed length of cloth, for example. Doing so made commerce possible, leading to improved communications and the spread of civilization.

Three to four thousand years ago, Egyptians introduced fractions. They first used them to show reciprocals. Later, they used them to represent the amount when a quantity was divided into equal parts.

But what if there were no cattle to trade or an entire crop of grain was lost in a flood? How could someone indicate the existence of nothing? From earliest times, people had thought of a “base state” while counting and used various symbols to represent this null condition. However, it was not until about the fifth century A.D. in India that zero was added to the number system and used as a numeral in calculations.

Clearly, there was also a need for numbers to represent loss or debt. In India, in the seventh century A.D., negative numbers were used as solutions to mathematical equations and commercial debts. The opposites of the counting numbers expanded the number system even further.

Because of the evolution of the number system, we can now perform complex calculations using these and other categories of real numbers. In this section, we will explore sets of numbers, calculations with different kinds of numbers, and the use of numbers in expressions.

The numbers we use for counting, or enumerating items, are the natural numbers: 1, 2, 3, 4, 5, and so on. We describe them in set notation as {1, 2, 3, …} where the ellipsis (…) indicates that the numbers continue to infinity. The natural numbers are, of course, also called the counting numbers. Any time we enumerate the members of a team, count the coins in a collection, or tally the trees in a grove, we are using the set of natural numbers. The set of whole numbers is the set of natural numbers plus zero: {0, 1, 2, 3,…}.

The set of integers adds the opposites of the natural numbers to the set of whole numbers: {…-3, -2, -1, 0, 1, 2, 3,…}. It is useful to note that the set of integers is made up of three distinct subsets: negative integers, zero, and positive integers. In this sense, the positive integers are just the natural numbers. Another way to think about it is that the natural numbers are a subset of the integers.

The set of rational numbers is written as [latex]\left\{\frac{m}{n}|m\text{ and }{n}\text{ are integers and }{n}\ne{ 0 }\right\}[/latex]. Notice from the definition that rational numbers are fractions (or quotients) containing integers in both the numerator and the denominator, and the denominator is never 0. We can also see that every natural number, whole number, and integer is a rational number with a denominator of 1.

Because they are fractions, any rational number can also be expressed in decimal form. Any rational number can be represented as either:

1. a terminating decimal: [latex]\frac{15}{8}=1.875\\[/latex], or
2. a repeating decimal: [latex]\frac{4}{11}=0.36363636\dots =0.\overline{36}\\[/latex]

We use a line drawn over the repeating block of numbers instead of writing the group multiple times.

Irrational Numbers

At some point in the ancient past, someone discovered that not all numbers are rational numbers. A builder, for instance, may have found that the diagonal of a square with unit sides was not 2 or even [latex]\frac{3}{2}\\[/latex], but was something else. Or a garment maker might have observed that the ratio of the circumference to the diameter of a roll of cloth was a little bit more than 3, but still not a rational number. Such numbers are said to be irrational because they cannot be written as fractions. These numbers make up the set of irrational numbers. Irrational numbers cannot be expressed as a fraction of two integers. It is impossible to describe this set of numbers by a single rule except to say that a number is irrational if it is not rational. So we write this as shown.

{h | h is not a rational number}

Example 3: Differentiating Rational and Irrational Numbers

Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.

1. [latex]\sqrt{25}\\[/latex]
2. [latex]\frac{33}{9}\\[/latex]
3. [latex]\sqrt{11}\\[/latex]
4. [latex]\frac{17}{34}\\[/latex]
5. [latex]0.3033033303333\dots\\[/latex]

Solution

1. [latex]\sqrt{25}:\\[/latex] This can be simplified as [latex]\sqrt{25}=5[/latex]. Therefore, [latex]\sqrt{25}\\[/latex] is rational.
2. [latex]\frac{33}{9}:\\[/latex] Because it is a fraction, [latex]\frac{33}{9}\\[/latex] is a rational number. Next, simplify and divide.
   [latex]\frac{33}{9}=\frac{{{11}\cdot{3}}}{{{3}\cot{3}}}=\frac{11}{3}=3.\overline{6}\\[/latex]

   So, [latex]\frac{33}{9}\\[/latex] is rational and a repeating decimal.

3. [latex]\sqrt{11}:\\[/latex] This cannot be simplified any further. Therefore, [latex]\sqrt{11}\\[/latex] is an irrational number.
4. [latex]\frac{17}{34}:\\[/latex] Because it is a fraction, [latex]\frac{17}{34}\\[/latex] is a rational number. Simplify and divide.
   [latex]\frac{17}{34}=\frac{{1}{\overline{)17}}}{\underset{2}{\overline{)34}}}=\frac{1}{2}=0.5\\[/latex]

   So, [latex]\frac{17}{34}\\[/latex] is rational and a terminating decimal.

5. 0.3033033303333… is not a terminating decimal. Also note that there is no repeating pattern because the group of 3s increases each time. Therefore it is neither a terminating nor a repeating decimal and, hence, not a rational number. It is an irrational number.

Try It 3

Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.

a. [latex]\frac{7}{77}\\[/latex]
b. [latex]\sqrt{81}\\[/latex]
c. [latex]4.27027002700027\dots\\[/latex]
d. [latex]\frac{91}{13}\\[/latex]
e. [latex]\sqrt{39}\\[/latex]

Solution

Real Numbers

Given any number n, we know that n is either rational or irrational. It cannot be both. The sets of rational and irrational numbers together make up the set of real numbers. As we saw with integers, the real numbers can be divided into three subsets: negative real numbers, zero, and positive real numbers. Each subset includes fractions, decimals, and irrational numbers according to their algebraic sign (+ or –). Zero is considered neither positive nor negative.

The real numbers can be visualized on a horizontal number line with an arbitrary point chosen as 0, with negative numbers to the left of 0 and positive numbers to the right of 0. A fixed unit distance is then used to mark off each integer (or other basic value) on either side of 0. Any real number corresponds to a unique position on the number line.The converse is also true: Each location on the number line corresponds to exactly one real number. This is known as a one-to-one correspondence. We refer to this as the real number line as shown in Figure 1.

Figure 1. The real number line

 

Example 4: Classifying Real Numbers

Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line?

1. [latex]-\frac{10}{3}\\[/latex]
2. [latex]\sqrt{5}\\[/latex]
3. [latex]-\sqrt{289}\\[/latex]
4. [latex]-6\pi\\[/latex]
5. [latex]0.615384615384\dots\\[/latex]

Solution

1. [latex]-\frac{10}{3}\\[/latex] is negative and rational. It lies to the left of 0 on the number line.
2. [latex]\sqrt{5}[/latex] is positive and irrational. It lies to the right of 0.
3. [latex]-\sqrt{289}=-\sqrt{{17}^{2}}=-17[/latex] is negative and rational. It lies to the left of 0.
4. [latex]-6\pi [/latex] is negative and irrational. It lies to the left of 0.
5. [latex]0.615384615384\dots [/latex] is a repeating decimal so it is rational and positive. It lies to the right of 0.

Try It 4

Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line?

a. [latex]\sqrt{73}\\[/latex]
b. [latex]-11.411411411\dots [/latex]
c. [latex]\frac{47}{19}[/latex]
d. [latex]-\frac{\sqrt{5}}{2}[/latex]
e. [latex]6.210735[/latex]

Solution

Sets of Numbers as Subsets

Beginning with the natural numbers, we have expanded each set to form a larger set, meaning that there is a subset relationship between the sets of numbers we have encountered so far. These relationships become more obvious when seen as a diagram.

 

Figure 2. Sets of numbers.   N: the set of natural numbers   W: the set of whole numbers   I: the set of integers   Q: the set of rational numbers   Q´: the set of irrational numbers

A General Note: Sets of Numbers

The set of natural numbers includes the numbers used for counting: [latex]\{1,2,3,\dots\}[/latex].

The set of whole numbers is the set of natural numbers plus zero: [latex]\{0,1,2,3,\dots\}[/latex].

The set of integers adds the negative natural numbers to the set of whole numbers: [latex]\{\dots,-3,-2,-1,0,1,2,3,\dots\}[/latex].

The set of rational numbers includes fractions written as [latex]\{\frac{m}{n}|m\text{ and }n\text{ are integers and }n\ne 0\}[/latex].

The set of irrational numbers is the set of numbers that are not rational, are nonrepeating, and are nonterminating: [latex]\{h|h\text{ is not a rational number}\}[/latex].

Example 5: Differentiating the Sets of Numbers

Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q’).

1. [latex]\sqrt{36}[/latex]
2. [latex]\frac{8}{3}\\[/latex]
3. [latex]\sqrt{73}\\[/latex]
4. [latex]-6[/latex]
5. [latex]3.2121121112\dots [/latex]

Solution

	N	W	I	Q	Q’
1. [latex]\sqrt{36}=6[/latex]	X	X	X	X
2. [latex]\frac{8}{3}=2.\overline{6}[/latex]				X
3. [latex]\sqrt{73}\\[/latex]					X
4. –6			X	X
5. [latex]3.2121121112\dots[/latex]					X

Try It 5

Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q’).

a. [latex]-\frac{35}{7}[/latex]
b. [latex]0[/latex]
c. [latex]\sqrt{169}[/latex]
d. [latex]\sqrt{24}[/latex]
e. [latex]4.763763763\dots [/latex]

Solution

When we multiply a number by itself, we square it or raise it to a power of 2. For example, [latex]{4}^{2}=4\cdot 4=16[/latex]. We can raise any number to any power. In general, the exponential notation [latex]{a}^{n}[/latex] means that the number or variable [latex]a[/latex] is used as a factor [latex]n[/latex] times.

In this notation, [latex]{a}^{n}[/latex] is read as the nth power of [latex]a[/latex], where [latex]a[/latex] is called the base and [latex]n[/latex] is called the exponent. A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. For example, [latex]24+6\cdot \frac{2}{3}-{4}^{2}[/latex] is a mathematical expression.

To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations. This is a sequence of rules for evaluating such expressions.

Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols.

The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right.

Let’s take a look at the expression provided.

There are no grouping symbols, so we move on to exponents or radicals. The number 4 is raised to a power of 2, so simplify [latex]{4}^{2}[/latex] as 16.

Next, perform multiplication or division, left to right.

Lastly, perform addition or subtraction, left to right.

Therefore, [latex]24+6\cdot \frac{2}{3}-{4}^{2}=12[/latex].

For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result.

For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics.

Commutative Properties

The commutative property of addition states that numbers may be added in any order without affecting the sum.

We can better see this relationship when using real numbers.

Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product.

Again, consider an example with real numbers.

It is important to note that neither subtraction nor division is commutative. For example, [latex]17 - 5[/latex] is not the same as [latex]5 - 17[/latex]. Similarly, [latex]20\div 5\ne 5\div 20[/latex].

Associative Properties

The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same.

Consider this example.

The associative property of addition tells us that numbers may be grouped differently without affecting the sum.

This property can be especially helpful when dealing with negative integers. Consider this example.

Are subtraction and division associative? Review these examples.

As we can see, neither subtraction nor division is associative.

Distributive Property

The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum.

This property combines both addition and multiplication (and is the only property to do so). Let us consider an example.

Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by –7, and adding the products.

To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example.

Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction.

A special case of the distributive property occurs when a sum of terms is subtracted.

For example, consider the difference [latex]12-\left(5+3\right)[/latex]. We can rewrite the difference of the two terms 12 and [latex]\left(5+3\right)[/latex] by turning the subtraction expression into addition of the opposite. So instead of subtracting [latex]\left(5+3\right)[/latex], we add the opposite.

Now, distribute [latex]-1[/latex] and simplify the result.

This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign of each term and add the results. With this in mind, we can rewrite the last example.

Identity Properties

The identity property of addition states that there is a unique number, called the additive identity (0) that, when added to a number, results in the original number.

The identity property of multiplication states that there is a unique number, called the multiplicative identity (1) that, when multiplied by a number, results in the original number.

For example, we have [latex]\left(-6\right)+0=-6[/latex] and [latex]23\cdot 1=23[/latex]. There are no exceptions for these properties; they work for every real number, including 0 and 1.

Inverse Properties

The inverse property of addition states that, for every real number a, there is a unique number, called the additive inverse (or opposite), denoted−a, that, when added to the original number, results in the additive identity, 0.

For example, if [latex]a=-8[/latex], the additive inverse is 8, since [latex]\left(-8\right)+8=0[/latex].

The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not defined. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted [latex]\frac{1}{a}[/latex], that, when multiplied by the original number, results in the multiplicative identity, 1.

For example, if [latex]a=-\frac{2}{3}[/latex], the reciprocal, denoted [latex]\frac{1}{a}[/latex], is [latex]-\frac{3}{2}[/latex] because

So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as [latex]x+5,\frac{4}{3}\pi {r}^{3}[/latex], or [latex]\sqrt{2{m}^{3}{n}^{2}}[/latex]. In the expression [latex]x+5[/latex], 5 is called a constant because it does not vary and x is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division.

We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way.

In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables.

Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before.

An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation [latex]2x+1=7[/latex] has the unique solution [latex]x=3[/latex] because when we substitute 3 for [latex]x[/latex] in the equation, we obtain the true statement [latex]2\left(3\right)+1=7[/latex].

A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area [latex]A[/latex] of a circle in terms of the radius [latex]r[/latex] of the circle: [latex]A=\pi {r}^{2}[/latex]. For any value of [latex]r[/latex], the area [latex]A[/latex] can be found by evaluating the expression [latex]\pi {r}^{2}[/latex].

Example 11: Using a Formula

A right circular cylinder with radius [latex]r[/latex] and height [latex]h[/latex] has the surface area [latex]S[/latex] (in square units) given by the formula [latex]S=2\pi r\left(r+h\right)[/latex]. Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of [latex]\pi[/latex].

Figure 3. Right circular cylinder

Try It 11

Figure 4

A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm²) is found to be [latex]A=\left(L+16\right)\left(W+16\right)-L\cdot W[/latex]. Find the area of a matte for a photograph with length 32 cm and width 24 cm.

Solution

Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions.

Key Concepts

• Rational numbers may be written as fractions or terminating or repeating decimals.
• Determine whether a number is rational or irrational by writing it as a decimal.
• The rational numbers and irrational numbers make up the set of real numbers. A number can be classified as natural, whole, integer, rational, or irrational.
• The order of operations is used to evaluate expressions.
• The real numbers under the operations of addition and multiplication obey basic rules, known as the properties of real numbers. These are the commutative properties, the associative properties, the distributive property, the identity properties, and the inverse properties.
• Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division. They take on a numerical value when evaluated by replacing variables with constants.
• Formulas are equations in which one quantity is represented in terms of other quantities. They may be simplified or evaluated as any mathematical expression.

Glossary

algebraic expression constants and variables combined using addition, subtraction, multiplication, and division

associative property of addition the sum of three numbers may be grouped differently without affecting the result; in symbols, [latex]a+\left(b+c\right)=\left(a+b\right)+c[/latex]

associative property of multiplication the product of three numbers may be grouped differently without affecting the result; in symbols, [latex]a\cdot \left(b\cdot c\right)=\left(a\cdot b\right)\cdot c[/latex]

base in exponential notation, the expression that is being multiplied

commutative property of addition two numbers may be added in either order without affecting the result; in symbols, [latex]a+b=b+a[/latex]

commutative property of multiplication two numbers may be multiplied in any order without affecting the result; in symbols, [latex]a\cdot b=b\cdot a[/latex]

constant a quantity that does not change value

distributive property the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols, [latex]a\cdot \left(b+c\right)=a\cdot b+a\cdot c[/latex]

equation a mathematical statement indicating that two expressions are equal

exponent in exponential notation, the raised number or variable that indicates how many times the base is being multiplied

exponential notation a shorthand method of writing products of the same factor

formula an equation expressing a relationship between constant and variable quantities

identity property of addition there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols, [latex]a+0=a[/latex]

identity property of multiplication there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols, [latex]a\cdot 1=a[/latex]

integers the set consisting of the natural numbers, their opposites, and 0: [latex]\{\dots ,-3,-2,-1,0,1,2,3,\dots \}[/latex]

inverse property of addition for every real number [latex]a[/latex], there is a unique number, called the additive inverse (or opposite), denoted [latex]-a[/latex], which, when added to the original number, results in the additive identity, 0; in symbols, [latex]a+\left(-a\right)=0[/latex]

inverse property of multiplication for every non-zero real number [latex]a[/latex], there is a unique number, called the multiplicative inverse (or reciprocal), denoted [latex]\frac{1}{a}[/latex], which, when multiplied by the original number, results in the multiplicative identity, 1; in symbols, [latex]a\cdot \frac{1}{a}=1[/latex]

irrational numbers the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot be expressed as a fraction of two integers

natural numbers the set of counting numbers: [latex]\{1,2,3,\dots \}[/latex]

order of operations a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to operations

rational numbers the set of all numbers of the form [latex]\frac{m}{n}[/latex], where [latex]m[/latex] and [latex]n[/latex] are integers and [latex]n\ne 0[/latex]. Any rational number may be written as a fraction or a terminating or repeating decimal.

real number line a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left.

real numbers the sets of rational numbers and irrational numbers taken together

variable a quantity that may change value

whole numbers the set consisting of 0 plus the natural numbers: [latex]\{0,1,2,3,\dots \}[/latex]

1. Is [latex]\sqrt{2}[/latex] an example of a rational terminating, rational repeating, or irrational number? Tell why it fits that category.

2. What is the order of operations? What acronym is used to describe the order of operations, and what does it stand for?

3. What do the Associative Properties allow us to do when following the order of operations? Explain your answer.

For the following exercises, simplify the given expression.

4. [latex]10+2\times \left(5 - 3\right)[/latex]

5. [latex]6\div 2-\left(81\div {3}^{2}\right)[/latex]

6. [latex]18+{\left(6 - 8\right)}^{3}[/latex]

7. [latex]-2\times {\left[16\div {\left(8 - 4\right)}^{2}\right]}^{2}[/latex]

8. [latex]4 - 6+2\times 7[/latex]

9. [latex]3\left(5 - 8\right)[/latex]

10. [latex]4+6 - 10\div 2[/latex]

11. [latex]12\div \left(36\div 9\right)+6[/latex]

12. [latex]{\left(4+5\right)}^{2}\div 3[/latex]

13. [latex]3 - 12\times 2+19[/latex]

14. [latex]2+8\times 7\div 4[/latex]

15. [latex]5+\left(6+4\right)-11[/latex]

16. [latex]9 - 18\div {3}^{2}[/latex]

17. [latex]14\times 3\div 7 - 6[/latex]

18. [latex]9-\left(3+11\right)\times 2[/latex]

19. [latex]6+2\times 2 - 1[/latex]

20. [latex]64\div \left(8+4\times 2\right)[/latex]

21. [latex]9+4\left({2}^{2}\right)[/latex]

22. [latex]{\left(12\div 3\times 3\right)}^{2}[/latex]

23. [latex]25\div {5}^{2}-7[/latex]

24. [latex]\left(15 - 7\right)\times \left(3 - 7\right)[/latex]

25. [latex]2\times 4 - 9\left(-1\right)[/latex]

26. [latex]{4}^{2}-25\times \frac{1}{5}[/latex]

27. [latex]12\left(3 - 1\right)\div 6[/latex]

For the following exercises, solve for the variable.

28. [latex]8\left(x+3\right)=64[/latex]

29. [latex]4y+8=2y[/latex]

30. [latex]\left(11a+3\right)-18a=-4[/latex]

31. [latex]4z - 2z\left(1+4\right)=36[/latex]

32. [latex]4y{\left(7 - 2\right)}^{2}=-200[/latex]

33. [latex]-{\left(2x\right)}^{2}+1=-3[/latex]

34. [latex]8\left(2+4\right)-15b=b[/latex]

35. [latex]2\left(11c - 4\right)=36[/latex]

36. [latex]4\left(3 - 1\right)x=4[/latex]

37. [latex]\frac{1}{4}\left(8w-{4}^{2}\right)=0[/latex]

For the following exercises, simplify the expression.

38. [latex]4x+x\left(13 - 7\right)[/latex]

39. [latex]2y-{\left(4\right)}^{2}y - 11[/latex]

40. [latex]\frac{a}{{2}^{3}}\left(64\right)-12a\div 6[/latex]

41. [latex]8b - 4b\left(3\right)+1[/latex]

42. [latex]5l\div 3l\times \left(9 - 6\right)[/latex]

43. [latex]7z - 3+z\times {6}^{2}[/latex]

44. [latex]4\times 3+18x\div 9 - 12[/latex]

45. [latex]9\left(y+8\right)-27[/latex]

46. [latex]\left(\frac{9}{6}t - 4\right)2[/latex]

47. [latex]6+12b - 3\times 6b[/latex]

48. [latex]18y - 2\left(1+7y\right)[/latex]

49. [latex]{\left(\frac{4}{9}\right)}^{2}\times 27x[/latex]

50. [latex]8\left(3-m\right)+1\left(-8\right)[/latex]

51. [latex]9x+4x\left(2+3\right)-4\left(2x+3x\right)[/latex]

52. [latex]{5}^{2}-4\left(3x\right)[/latex]

For the following exercises, consider this scenario: Fred earns $40 mowing lawns. He spends $10 on mp3s, puts half of what is left in a savings account, and gets another $5 for washing his neighbor’s car.

53. Write the expression that represents the number of dollars Fred keeps (and does not put in his savings account). Remember the order of operations.

54. How much money does Fred keep?

For the following exercises, solve the given problem.

55. According to the U.S. Mint, the diameter of a quarter is 0.955 inches. The circumference of the quarter would be the diameter multiplied by [latex]\pi [/latex]. Is the circumference of a quarter a whole number, a rational number, or an irrational number?

56. Jessica and her roommate, Adriana, have decided to share a change jar for joint expenses. Jessica put her loose change in the jar first, and then Adriana put her change in the jar. We know that it does not matter in which order the change was added to the jar. What property of addition describes this fact?

For the following exercises, consider this scenario: There is a mound of [latex]g[/latex] pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel is added to the mound. Two orders of 600 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,200 pounds of gravel.

57. Write the equation that describes the situation.

58. Solve for g.

For the following exercise, solve the given problem.

59. Ramon runs the marketing department at his company. His department gets a budget every year, and every year, he must spend the entire budget without going over. If he spends less than the budget, then his department gets a smaller budget the following year. At the beginning of this year, Ramon got $2.5 million for the annual marketing budget. He must spend the budget such that [latex]2,500,000-x=0[/latex]. What property of addition tells us what the value of x must be?

For the following exercises, use a graphing calculator to solve for x. Round the answers to the nearest hundredth.

60. [latex]0.5{\left(12.3\right)}^{2}-48x=\frac{3}{5}[/latex]

61. [latex]{\left(0.25 - 0.75\right)}^{2}x - 7.2=9.9[/latex]

62. If a whole number is not a natural number, what must the number be?

63. Determine whether the statement is true or false: The multiplicative inverse of a rational number is also rational.

64. Determine whether the statement is true or false: The product of a rational and irrational number is always irrational.

65. Determine whether the simplified expression is rational or irrational: [latex]\sqrt{-18 - 4\left(5\right)\left(-1\right)}[/latex].

66. Determine whether the simplified expression is rational or irrational: [latex]\sqrt{-16+4\left(5\right)+5}[/latex].

67. The division of two whole numbers will always result in what type of number?

68. What property of real numbers would simplify the following expression: [latex]4+7\left(x - 1\right)?[/latex]

Solutions to Try Its

1. a. [latex]\frac{11}{1}[/latex]
b. [latex]\frac{3}{1}[/latex]
c. [latex]-\frac{4}{1}[/latex]

2. a. 4 (or 4.0), terminating
b. [latex]0.\overline{615384}[/latex], repeating
c. –0.85, terminating

3. a. rational and repeating
b. rational and terminating
c. irrational
d. rational and repeating
e. irrational

4. a. positive, irrational; right
b. negative, rational; left
c. positive, rational; right
d. negative, irrational; left
e. positive, rational; right

5.

6. a. 10
b. 2
c. 4.5
d. 25
e. 26

7. a. commutative property of multiplication, associative property of multiplication, inverse property of multiplication, identity property of multiplication;
b. 33, distributive property;
c. 26, distributive property;
d. [latex]\frac{4}{9}[/latex], commutative property of addition, associative property of addition, inverse property of addition, identity property of addition;
e. 0, distributive property, inverse property of addition, identity property of addition

8.

9. a. 5
b. 11
c. 9
d. 26

10. a. 4
b. 11
c. [latex]\frac{121}{3}\pi [/latex]
d. 1728
e. 3

11. 1,152 cm²

12. a. [latex]-2y - 2z\text{ or }-2\left(y+z\right)[/latex]
b. [latex]\frac{2}{t}-1[/latex]
c. [latex]3pq - 4p+q[/latex]
d. [latex]7r - 2s+6[/latex]

13. [latex]A=P\left(1+rt\right)[/latex]

Solutions to Odd-Numbered Exercises

1. irrational number. The square root of two does not terminate, and it does not repeat a pattern. It cannot be written as a quotient of two integers, so it is irrational.

3. The Associative Properties state that the sum or product of multiple numbers can be grouped differently without affecting the result. This is because the same operation is performed (either addition or subtraction), so the terms can be re-ordered.

5. [latex]-6[/latex]

7. [latex]-2[/latex]

9. [latex]-9[/latex]

11. 9

13. 4

15. 4

17. 0

19. 9

21. 25

23. [latex]-6[/latex]

25. 17

27. 4

29. [latex]-4[/latex]

31. [latex]-6[/latex]

33. [latex]\pm 1[/latex]

35. 2

37. 2

39. [latex]-14y - 11[/latex]

41. [latex]-4b+1[/latex]

43. [latex]43z - 3[/latex]

45. [latex]9y+45[/latex]

47. [latex]-6b+6[/latex]

49. [latex]\frac{16x}{3}\\[/latex]

51. [latex]9x[/latex]

53. [latex]\frac{1}{2}\left(40 - 10\right)+5[/latex]

55. irrational number

57. [latex]g+400 - 2\left(600\right)=1200[/latex]

59. inverse property of addition

61. 68.4

63. true

65. irrational

67. rational

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1.

Figure 1

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form [latex]ax+b=0[/latex] and are solved using basic algebraic operations.

We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.

The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for [latex]x[/latex] will make the equation true.

A conditional equation is true for only some values of the variable. For example, if we are to solve the equation [latex]5x+2=3x - 6[/latex], we have the following:

The solution set consists of one number: [latex]\{-4\}[/latex]. It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve [latex]5x - 15=5\left(x - 4\right)[/latex], we have the following:

Indeed, [latex]-15\ne -20[/latex]. There is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation.

Recall that a rational number is the ratio of two numbers, such as [latex]\frac{2}{3}[/latex] or [latex]\frac{7}{2}[/latex]. A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

Rational equations have a variable in the denominator in at least one of the terms.
Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as [latex]\left(x+1\right)[/latex]. Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are [latex]x[/latex], [latex]x - 1[/latex], and [latex]3x - 3[/latex]. First, factor all denominators. We then have [latex]x[/latex], [latex]\left(x - 1\right)[/latex], and [latex]3\left(x - 1\right)[/latex] as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of [latex]\left(x - 1\right)[/latex]. The [latex]x[/latex] in the first denominator is separate from the [latex]x[/latex] in the [latex]\left(x - 1\right)[/latex] denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the [latex]x[/latex], one factor of [latex]\left(x - 1\right)[/latex], and the 3. Thus, the LCD is the following:

So, both sides of the equation would be multiplied by [latex]3x\left(x - 1\right)[/latex]. Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as [latex]x[/latex] and [latex]{x}^{2}+2x[/latex]. Once the second denominator is factored as [latex]{x}^{2}+2x=x\left(x+2\right)[/latex], there is a common factor of x in both denominators and the LCD is [latex]x\left(x+2\right)[/latex].

Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.

Multiply [latex]a\left(d\right)[/latex] and [latex]b\left(c\right)[/latex], which results in [latex]ad=bc[/latex].

Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as [latex]y=mx+b[/latex], where [latex]m=\text{slope}[/latex] and [latex]b=y\text{-intercept}[/latex]. Let us begin with the slope.

The Slope of a Line

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2. The lines indicate the following slopes: [latex]m=-3[/latex], [latex]m=2[/latex], and [latex]m=\frac{1}{3}[/latex].

Figure 2

The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.

This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

Standard Form of a Line

Another way that we can represent the equation of a line is in standard form. Standard form is given as

where [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side.

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c.

Suppose that we want to find the equation of a line containing the following points: [latex]\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)[/latex], and [latex]\left(-3,5\right)[/latex]. First, we will find the slope.

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through [latex]x=-3[/latex].

The equation of a horizontal line is given as

where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c.

Suppose we want to find the equation of a line that contains the following set of points: [latex]\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)[/latex], and [latex]\left(5,-2\right)[/latex]. We can use the point-slope formula. First, we find the slope using any two points on the line.

Use any point for [latex]\left({x}_{1},{y}_{1}\right)[/latex] in the formula, or use the y-intercept.

The graph is a horizontal line through [latex]y=-2[/latex]. Notice that all of the y-coordinates are the same.

Figure 3. The line x = −3 is a vertical line. The line y = −2 is a horizontal line.

Figure 4. Parallel lines

Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, [latex]m=2[/latex].

All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts.

Lines that are perpendicular intersect to form a [latex]{90}^{\circ }[/latex] -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\cdot {m}_{2}=-1[/latex]. For example, Figure 4 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\frac{1}{3}[/latex].

Figure 5. Perpendicular lines

Solution

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

[latex]\begin{array}{l}3y=-4x+3\hfill \\ y=-\frac{4}{3}x+1\hfill \end{array}[/latex]

Second equation:

[latex]\begin{array}{l}3x - 4y=8\hfill \\ -4y=-3x+8\hfill \\ y=\frac{3}{4}x - 2\hfill \end{array}[/latex]

See the graph of both lines in Figure 6.

Figure 6

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

[latex]\begin{array}{l}{m}_{1}=-\frac{4}{3}\hfill \\ {m}_{2}=\frac{3}{4}\hfill \\ {m}_{1}\cdot {m}_{2}=\left(-\frac{4}{3}\right)\left(\frac{3}{4}\right)=-1\hfill \end{array}[/latex]

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Solution

First, we will write the equation in slope-intercept form to find the slope.

[latex]\begin{array}{l}5x+3y=1\hfill \\ 3y=-5x+1\hfill \\ y=-\frac{5}{3}x+\frac{1}{3}\hfill \end{array}[/latex]

The slope is [latex]m=-\frac{5}{3}[/latex]. The y-intercept is [latex]\frac{1}{3}[/latex], but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

[latex]\begin{array}{l}y - 5=-\frac{5}{3}\left(x - 3\right)\hfill \\ y - 5=-\frac{5}{3}x+5\hfill \\ y=-\frac{5}{3}x+10\hfill \end{array}[/latex]

The equation of the line is [latex]y=-\frac{5}{3}x+10[/latex].

Figure 7

Key Concepts

• We can solve linear equations in one variable in the form [latex]ax+b=0[/latex] using standard algebraic properties.
• A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation.
• All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator.
• Given two points, we can find the slope of a line using the slope formula.
• We can identify the slope and y-intercept of an equation in slope-intercept form.
• We can find the equation of a line given the slope and a point.
• We can also find the equation of a line given two points. Find the slope and use the point-slope formula.
• The standard form of a line has no fractions.
• Horizontal lines have a slope of zero and are defined as [latex]y=c[/latex], where c is a constant.
• Vertical lines have an undefined slope (zero in the denominator), and are defined as [latex]x=c[/latex], where c is a constant.
• Parallel lines have the same slope and different y-intercepts.
• Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical.

Glossary

conditional equation an equation that is true for some values of the variable

identity equation an equation that is true for all values of the variable

inconsistent equation an equation producing a false result

linear equation an algebraic equation in which each term is either a constant or the product of a constant and the first power of a variable

solution set the set of all solutions to an equation

slope the change in y-values over the change in x-values

rational equation an equation consisting of a fraction of polynomials

1. What does it mean when we say that two lines are parallel?

2. What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

3. How do we recognize when an equation, for example [latex]y=4x+3[/latex], will be a straight line (linear) when graphed?

4. What does it mean when we say that a linear equation is inconsistent?

5. When solving the following equation: [latex]\frac{2}{x - 5}=\frac{4}{x+1}[/latex], explain why we must exclude [latex]x=5[/latex] and [latex]x=-1[/latex] as possible solutions from the solution set.

For the following exercises, solve the equation for [latex]x[/latex].

6. [latex]7x+2=3x - 9[/latex]

7. [latex]4x - 3=5[/latex]

8. [latex]3\left(x+2\right)-12=5\left(x+1\right)[/latex]

9. [latex]12 - 5\left(x+3\right)=2x - 5[/latex]

10. [latex]\frac{1}{2}-\frac{1}{3}x=\frac{4}{3}[/latex]

11. [latex]\frac{x}{3}-\frac{3}{4}=\frac{2x+3}{12}[/latex]

12. [latex]\frac{2}{3}x+\frac{1}{2}=\frac{31}{6}[/latex]

13. [latex]3\left(2x - 1\right)+x=5x+3[/latex]

14. [latex]\frac{2x}{3}-\frac{3}{4}=\frac{x}{6}+\frac{21}{4}[/latex]

15. [latex]\frac{x+2}{4}-\frac{x - 1}{3}=2[/latex]

For the following exercises, solve each rational equation for [latex]x[/latex]. State all x-values that are excluded from the solution set.

16. [latex]\frac{3}{x}-\frac{1}{3}=\frac{1}{6}[/latex]

17. [latex]2-\frac{3}{x+4}=\frac{x+2}{x+4}[/latex]

18. [latex]\frac{3}{x - 2}=\frac{1}{x - 1}+\frac{7}{\left(x - 1\right)\left(x - 2\right)}[/latex]

19. [latex]\frac{3x}{x - 1}+2=\frac{3}{x - 1}[/latex]

20. [latex]\frac{5}{x+1}+\frac{1}{x - 3}=\frac{-6}{{x}^{2}-2x - 3}[/latex]

21. [latex]\frac{1}{x}=\frac{1}{5}+\frac{3}{2x}[/latex]

For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.

22. [latex]\left(0,3\right)[/latex] with a slope of [latex]\frac{2}{3}[/latex]

23. [latex]\left(1,2\right)[/latex] with a slope of [latex]\frac{-4}{5}[/latex]

24. x-intercept is 1, and [latex]\left(-2,6\right)[/latex]

25. y-intercept is 2, and [latex]\left(4,-1\right)[/latex]

26. [latex]\left(-3,10\right)[/latex] and [latex]\left(5,-6\right)[/latex]

27. [latex]\left(1,3\right)\text{ and }\left(5,5\right)[/latex]

28. parallel to [latex]y=2x+5[/latex] and passes through the point [latex]\left(4,3\right)[/latex]

29. perpendicular to [latex]\text{3}y=x - 4[/latex] and passes through the point [latex]\left(-2,1\right)[/latex] .

For the following exercises, find the equation of the line using the given information.

30. [latex]\left(-2,0\right)[/latex] and [latex]\left(-2,5\right)[/latex]

31. [latex]\left(1,7\right)[/latex] and [latex]\left(3,7\right)[/latex]

32. The slope is undefined and it passes through the point [latex]\left(2,3\right)[/latex].

33. The slope equals zero and it passes through the point [latex]\left(1,-4\right)[/latex].

34. The slope is [latex]\frac{3}{4}[/latex] and it passes through the point [latex]\left(1,4\right)[/latex].

35. [latex]\left(-1,3\right)[/latex] and [latex]\left(4,-5\right)[/latex]

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

36. [latex]\begin{array}{l}\\ y=2x+7\hfill \\ y=\frac{-1}{2}x - 4\hfill \end{array}[/latex]

37. [latex]\begin{array}{l}3x - 2y=5\hfill \\ 6y - 9x=6\hfill \end{array}[/latex]

38. [latex]\begin{array}{l}y=\frac{3x+1}{4}\hfill \\ y=3x+2\hfill \end{array}[/latex]

39. [latex]\begin{array}{l}x=4\\ y=-3\end{array}[/latex]

For the following exercises, find the slope of the line that passes through the given points.

40. [latex]\left(5,4\right)[/latex] and [latex]\left(7,9\right)[/latex]

41. [latex]\left(-3,2\right)[/latex] and [latex]\left(4,-7\right)[/latex]

42. [latex]\left(-5,4\right)[/latex] and [latex]\left(2,4\right)[/latex]

43. [latex]\left(-1,-2\right)[/latex] and [latex]\left(3,4\right)[/latex]

44. [latex]\left(3,-2\right)[/latex] and [latex]\left(3,-2\right)[/latex]

For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

45. [latex]\begin{array}{l}\left(-1,3\right)\text{ and }\left(5,1\right)\\ \left(-2,3\right)\text{ and }\left(0,9\right)\end{array}[/latex]

46. [latex]\begin{array}{l}\left(2,5\right)\text{ and }\left(5,9\right)\\ \left(-1,-1\right)\text{ and }\left(2,3\right)\end{array}[/latex]

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y-intercept occurs. State your ymin and ymax values.

47. [latex]0.537x - 2.19y=100[/latex]

48. [latex]4,500x - 200y=9,528[/latex]

49. [latex]\frac{200 - 30y}{x}=70[/latex]

50. Starting with the point-slope formula [latex]y-{y}_{1}=m\left(x-{x}_{1}\right)[/latex], solve this expression for [latex]x[/latex] in terms of [latex]{x}_{1},y,{y}_{1}[/latex], and [latex]m[/latex].

51. Starting with the standard form of an equation [latex]\text{A}x\text{ + B}y\text{ = C,}[/latex] solve this expression for y in terms of [latex]A,B,C[/latex], and [latex]x[/latex]. Then put the expression in slope-intercept form.

52. Use the above derived formula to put the following standard equation in slope intercept form: [latex]7x - 5y=25[/latex].

53. Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.

54. Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

55. The slope for a wheelchair ramp for a home has to be [latex]\frac{1}{12}[/latex]. If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.

56. If the profit equation for a small business selling [latex]x[/latex] number of item one and [latex]y[/latex] number of item two is [latex]p=3x+4y[/latex], find the [latex]y[/latex] value when [latex]p=\$453\text{ and }x=75[/latex].

For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be [latex]y=45+.25x[/latex], where [latex]x[/latex] is the number of miles traveled.

57. What is your cost if you travel 50 mi?

58. If your cost were [latex]\$63.75[/latex], how many miles were you charged for traveling?

59. Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

Solutions to Try Its

1. [latex]x=-5[/latex]

2. [latex]x=-3[/latex]

3. [latex]x=\frac{10}{3}[/latex]

4. [latex]x=1[/latex]

5. [latex]x=-\frac{7}{17}[/latex]. Excluded values are [latex]x=-\frac{1}{2}[/latex] and [latex]x=-\frac{1}{3}[/latex].

6. [latex]x=\frac{1}{3}[/latex]

7. [latex]m=-\frac{2}{3}[/latex]

8. [latex]y=4x - 3[/latex]

9. [latex]x+3y=2[/latex]

10. Horizontal line: [latex]y=2[/latex]

11. Parallel lines: equations are written in slope-intercept form.

12. [latex]y=5x+3[/latex]

Solutions to Odd-Numbered Exercises

1. It means they have the same slope.

3. The exponent of the [latex]x[/latex] variable is 1. It is called a first-degree equation.

5. If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

7. [latex]x=2[/latex]

9. [latex]x=\frac{2}{7}[/latex]

11. [latex]x=6[/latex]

13. [latex]x=3[/latex]

15. [latex]x=-14[/latex]

17. [latex]x\ne -4[/latex]; [latex]x=-3[/latex]

19. [latex]x\ne 1[/latex]; when we solve this we get [latex]x=1[/latex], which is excluded, therefore NO solution

21. [latex]x\ne 0[/latex]; [latex]x=\frac{-5}{2}[/latex]

23. [latex]y=\frac{-4}{5}x+\frac{14}{5}[/latex]

25. [latex]y=\frac{-3}{4}x+2[/latex]

27. [latex]y=\frac{1}{2}x+\frac{5}{2}[/latex]

29. [latex]y=-3x - 5[/latex]

31. [latex]y=7[/latex]

33. [latex]y=-4[/latex]

35. [latex]8x+5y=7[/latex]

37. Parallel

39. Perpendicular

41. [latex]m=\frac{-9}{7}[/latex]

43. [latex]m=\frac{3}{2}[/latex]

45. [latex]{m}_{1}=\frac{-1}{3},\text{ }{m}_{2}=3;\text{ }\text{Perpendicular}[/latex]

47. [latex]y=0.245x - 45.662[/latex]. Answers may vary. [latex]{y}_{\text{min}}=-50,\text{ }{y}_{\text{max}}=-40[/latex]

49. [latex]y=-2.333x+6.667[/latex]. Answers may vary. [latex]{y}_{\mathrm{min}}=-10, {y}_{\mathrm{max}}=10[/latex]

51. [latex]y=\frac{-A}{B}x+\frac{C}{B}[/latex]

53. Yes, they are perpendicular.

55. 30 ft

57. $57.50

59. 220 mi

Figure 1. Credit: Kevin Dooley

Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[/latex]. This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost [latex]C[/latex].

When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table lists some common verbal expressions and their equivalent mathematical expressions.

Solution

1. The model for Company A can be written as [latex]A=0.05x+34[/latex]. This includes the variable cost of [latex]0.05x[/latex] plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[/latex]. Company B’s model can be written as [latex]B=0.04x+\$40[/latex].
2. If the average number of minutes used each month is 1,160, we have the following:
   [latex]\begin{array}{l}\text{Company }A\hfill&=0.05\left(1,160\right)+34\hfill \\ \hfill&=58+34\hfill \\ \hfill&=92\hfill \\ \hfill \\ \text{Company }B\hfill&=0.04\left(1,160\right)+40\hfill \\ \hfill&=46.4+40\hfill \\ \hfill&=86.4\hfill \end{array}[/latex]

   So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.

3. If the average number of minutes used each month is 420, we have the following:
   [latex]\begin{array}{l}\text{Company }A\hfill&=0.05\left(420\right)+34\hfill \\ \hfill&=21+34\hfill \\ \hfill&=55\hfill \\ \hfill \\ \text{Company }B\hfill&=0.04\left(420\right)+40\hfill \\ \hfill&=16.8+40\hfill \\ \hfill&=56.8\hfill \end{array}[/latex]

   If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of $56.80.

4. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of [latex]\left(x,y\right)[/latex] coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x.
   [latex]\begin{array}{l}0.05x+34=0.04x+40\hfill \\ 0.01x=6\hfill \\ x=600\hfill \end{array}[/latex]

   Check the x-value in each equation.

   [latex]\begin{array}{l}0.05\left(600\right)+34=64\hfill \\ 0.04\left(600\right)+40=64\hfill \end{array}[/latex]

   Therefore, a monthly average of 600 talk-time minutes renders the plans equal.

Figure 2

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, [latex]A=LW[/latex]; the perimeter of a rectangle, [latex]P=2L+2W[/latex]; and the volume of a rectangular solid, [latex]V=LWH[/latex]. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

Solution

The perimeter formula is standard: [latex]P=2L+2W[/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides.

Figure 3

Now we can solve for the width and then calculate the length.

[latex]\begin{array}{l}P=2L+2W\hfill \\ 54=2\left(W+3\right)+2W\hfill \\ 54=2W+6+2W\hfill \\ 54=4W+6\hfill \\ 48=4W\hfill \\ 12=W\hfill \\ \left(12+3\right)=L\hfill \\ 15=L\hfill \end{array}[/latex]

The dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.

Key Concepts

• A linear equation can be used to solve for an unknown in a number problem.
• Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities.
• There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the [latex]d=rt[/latex] formula.
• Many geometry problems are solved using the perimeter formula [latex]P=2L+2W[/latex], the area formula [latex]A=LW[/latex], or the volume formula [latex]V=LWH[/latex].

Glossary

area in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: [latex]A=LW[/latex]

perimeter in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: [latex]P=2L+2W[/latex]

volume in cubic units, the volume measurement includes length, width, and depth: [latex]V=LWH[/latex]

1. To set up a model linear equation to fit real-world applications, what should always be the first step?

2. Use your own words to describe this equation where n is a number:

3. If the total amount of money you had to invest was $2,000 and you deposit [latex]x[/latex] amount in one investment, how can you represent the remaining amount?

4. If a man sawed a 10-ft board into two sections and one section was [latex]n[/latex] ft long, how long would the other section be in terms of [latex]n[/latex] ?

5. If Bill was traveling [latex]v[/latex] mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster?

For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked.

6. Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?

7. Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages?

8. Ben originally filled out 8 more applications than Henry. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out?

For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls.

9. Find the model of the total cost of Company A’s plan, using [latex]m[/latex] for the minutes.

10. Find the model of the total cost of Company B’s plan, using [latex]m[/latex] for the minutes.

11. Find out how many minutes of calling would make the two plans equal.

12. If the person makes a monthly average of 200 min of calls, which plan should for the person choose?

For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 5 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use [latex]P[/latex] for the number of devices that need data plans as part of their cost.

13. Find the model of the total cost of the Family Plan.

14. Find the model of the total cost of the Mobile Share Plan.

15. Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost.

16. If a family has 3 smart phones, which plan should they choose?

For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%.

17. If we let [latex]x[/latex] be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD?

18. Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return.

19. Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart?

20. Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben?

21. Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h?

22. A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result?

23. Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000?

For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven.

24. Write the model equation for the cost of renting a truck with plan A.

25. Write the model equation for the cost of renting a truck with plan B.

26. Find the number of miles that would generate the same cost for both plans.

27. If Tim knows he has to travel 300 mi, which plan should he choose?

For the following exercises, use the given formulas to answer the questions.

28. [latex]A=P\left(1+rt\right)[/latex] is used to find the principal amount P deposited, earning r% interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if [latex]A=\$8,000[/latex].

29. The formula [latex]F=\frac{m{v}^{2}}{R}[/latex] relates force (F), velocity (v), mass (m), and resistance (R). Find [latex]R[/latex] when [latex]m=45[/latex], [latex]v=7[/latex], and [latex]F=245[/latex].

30. [latex]F=ma[/latex] indicates that force (F) equals mass (m) times acceleration (a). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it.

31. [latex]Sum=\frac{1}{1-r}[/latex] is the formula for an infinite series sum. If the sum is 5, find [latex]r[/latex].

For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question.

32. Solve for W: [latex]P=2L+2W[/latex]

33. Use the formula from the previous question to find the width, [latex]W[/latex], of a rectangle whose length is 15 and whose perimeter is 58.

34. Solve for [latex]f:\frac{1}{p}+\frac{1}{q}=\frac{1}{f}[/latex]

35. Use the formula from the previous question to find [latex]f[/latex] when [latex]p=8\text{and }q=13[/latex].

36. Solve for [latex]m[/latex] in the slope-intercept formula: [latex]y=mx+b[/latex]

37. Use the formula from the previous question to find [latex]m[/latex] when the coordinates of the point are [latex]\left(4,7\right)[/latex] and [latex]b=12[/latex].

38. The area of a trapezoid is given by [latex]A=\frac{1}{2}h\left({b}_{1}+{b}_{2}\right)[/latex]. Use the formula to find the area of a trapezoid with [latex]h=6,\text{ }{b}_{1}=14,\text{ and }{b}_{2}=8[/latex].

39. Solve for h: [latex]A=\frac{1}{2}h\left({b}_{1}+{b}_{2}\right)[/latex]

40. Use the formula from the previous question to find the height of a trapezoid with [latex]A=150,\text{ }{b}_{1}=19,\text{ and }{b}_{2}=11[/latex].

41. Find the dimensions of an American football field. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula [latex]P=2L+2W[/latex].

42. Distance equals rate times time, [latex]d=rt[/latex]. Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h.

43. Using the formula in the previous exercise, find the distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h.

44. What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h?

45. If the area model for a triangle is [latex]A=\frac{1}{2}bh[/latex], find the area of a triangle with a height of 16 in. and a base of 11 in.

46. Solve for h: [latex]A=\frac{1}{2}bh[/latex]

47. Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215.

48. The volume formula for a cylinder is [latex]V=\pi {r}^{2}h[/latex]. Using the symbol [latex]\pi [/latex] in your answer, find the volume of a cylinder with a radius, [latex]r[/latex], of 4 cm and a height of 14 cm.

49. Solve for h: [latex]V=\pi {r}^{2}h[/latex]

50. Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of [latex]16\pi [/latex]

51. Solve for r: [latex]V=\pi {r}^{2}h[/latex]

52. Use the formula from the previous question to find the radius of a cylinder with a height of 36 and a volume of [latex]324\pi [/latex].

53. The formula for the circumference of a circle is [latex]C=2\pi r[/latex]. Find the circumference of a circle with a diameter of 12 in. (diameter = 2r). Use the symbol [latex]\pi [/latex] in your final answer.

54. Solve the formula from the previous question for [latex]\pi [/latex]. Notice why [latex]\pi [/latex] is sometimes defined as the ratio of the circumference to its diameter.

Solutions to Try Its

1. 11 and 25

2. [latex]C=2.5x+3,650[/latex]

3. 45 mi/h

4. [latex]L=37[/latex] cm, [latex]W=18[/latex] cm

5. 250 ft²

Solutions to Odd-Numbered Exercises

1. Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

3. [latex]2,000-x[/latex]

5. [latex]v+10[/latex]

7. Ann: [latex]23[/latex]; Beth: [latex]46[/latex]

9. [latex]20+0.05m[/latex]

11. 300 min

13. [latex]90+40P[/latex]

15. 6 devices

17. [latex]50,000-x[/latex]

19. 4 h

21. She traveled for 2 h at 20 mi/h, or 40 miles.

23. $5,000 at 8% and $15,000 at 12%

25. [latex]B=100+.05x[/latex]

27. Plan A

29. [latex]R=9[/latex]

31. [latex]r=\frac{4}{5}[/latex] or 0.8

33. [latex]W=\frac{P - 2L}{2}=\frac{58 - 2\left(15\right)}{2}=14[/latex]

35. [latex]f=\frac{pq}{p+q}=\frac{8\left(13\right)}{8+13}=\frac{104}{21}[/latex]

37. [latex]m=\frac{-5}{4}[/latex]

39. [latex]h=\frac{2A}{{b}_{1}+{b}_{2}}[/latex]

41. length = 360 ft; width = 160 ft

43. 405 mi

45. [latex]A=88\text{ in}{.}^{2}[/latex]

47. 28.7

49. [latex]h=\frac{V}{\pi {r}^{2}}[/latex]

51. [latex]r=\sqrt{\frac{V}{\pi h}}[/latex]

53. [latex]C=12\pi [/latex]

The study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still have no solution to equations such as

Our best guesses might be +2 or –2. But if we test +2 in this equation, it does not work. If we test –2, it does not work. If we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides solutions to problems such as these. In this section, we will explore this number system and how to work within it.

We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a, b), where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis.

Figure 2

Let’s consider the number [latex]-2+3i\\[/latex]. The real part of the complex number is –2 and the imaginary part is 3i. We plot the ordered pair [latex]\left(-2,3\right)\\[/latex] to represent the complex number [latex]-2+3i\\[/latex].

A General Note: Complex Plane

Figure 3

In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis.

Example 2: Plotting a Complex Number on the Complex Plane

Plot the complex number [latex]3 - 4i\\[/latex] on the complex plane.

Solution

The real part of the complex number is 3, and the imaginary part is –4i. We plot the ordered pair [latex]\left(3,-4\right)\\[/latex].

Figure 4

Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts.

Watch this video online: https://youtu.be/SGhTjioGqqA

Multiplying Complex Numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

Example 4: Multiplying a Complex Number by a Real Number

Figure 5

Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

How To: Given a complex number and a real number, multiply to find the product.

1. Use the distributive property.
2. Simplify.

Example 5: Multiplying a Complex Number by a Real Number

Find the product [latex]4\left(2+5i\right)\\[/latex].

Solution

Distribute the 4.

[latex]\begin{cases}4\left(2+5i\right)=\left(4\cdot 2\right)+\left(4\cdot 5i\right)\hfill \\ =8+20i\hfill \end{cases}\\[/latex]

Try It 4

Find the product [latex]-4\left(2+6i\right)\\[/latex].

Solution

Multiplying Complex Numbers Together

Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get

[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci+bd{i}^{2}\\[/latex]

Because [latex]{i}^{2}=-1\\[/latex], we have

[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci-bd\\[/latex]

To simplify, we combine the real parts, and we combine the imaginary parts.

[latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i\\[/latex]

How To: Given two complex numbers, multiply to find the product.

1. Use the distributive property or the FOIL method.
2. Simplify.

Example 6: Multiplying a Complex Number by a Complex Number

Multiply [latex]\left(4+3i\right)\left(2 - 5i\right)\\[/latex].

Solution

Use [latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i\\[/latex]

[latex]\begin{cases}\left(4+3i\right)\left(2 - 5i\right)=\left(4\cdot 2 - 3\cdot \left(-5\right)\right)+\left(4\cdot \left(-5\right)+3\cdot 2\right)i\hfill \\ \text{ }=\left(8+15\right)+\left(-20+6\right)i\hfill \\ \text{ }=23 - 14i\hfill \end{cases}\\[/latex]

Try It 5

Multiply [latex]\left(3 - 4i\right)\left(2+3i\right)\\[/latex].

Solution

https://youtu.be/O9xQaIi0NX0

Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi\\[/latex] is [latex]a-bi\\[/latex].

Note that complex conjugates have a reciprocal relationship: The complex conjugate of [latex]a+bi\\[/latex] is [latex]a-bi\\[/latex], and the complex conjugate of [latex]a-bi\\[/latex] is [latex]a+bi\\[/latex]. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose we want to divide [latex]c+di\\[/latex] by [latex]a+bi\\[/latex], where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

Multiply the numerator and denominator by the complex conjugate of the denominator.

Apply the distributive property.

Simplify, remembering that [latex]{i}^{2}=-1\\[/latex].

Example 9: Substituting a Complex Number into a Polynomial Function

Let [latex]f\left(x\right)={x}^{2}-5x+2\\[/latex]. Evaluate [latex]f\left(3+i\right)\\[/latex].

Solution

Figure 6

Substitute [latex]x=3+i\\[/latex] into the function [latex]f\left(x\right)={x}^{2}-5x+2\\[/latex] and simplify.

Analysis of the Solution

We write [latex]f\left(3+i\right)=-5+i\\[/latex]. Notice that the input is [latex]3+i\\[/latex] and the output is [latex]-5+i\\[/latex].

Watch this video online: https://youtu.be/XBJjbJAwM1c

Simplifying Powers of i

The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.

We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.

Key Concepts

• The square root of any negative number can be written as a multiple of i.
• To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.
• Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts.
• Complex numbers can be multiplied and divided.
• To multiply complex numbers, distribute just as with polynomials.
• To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator.
• The powers of i are cyclic, repeating every fourth one.

Glossary

complex conjugate

the complex number in which the sign of the imaginary part is changed and the real part of the number is left unchanged; when added to or multiplied by the original complex number, the result is a real number

complex number

the sum of a real number and an imaginary number, written in the standard form a + bi, where a is the real part, and bi is the imaginary part

complex plane

a coordinate system in which the horizontal axis is used to represent the real part of a complex number and the vertical axis is used to represent the imaginary part of a complex number

imaginary number

a number in the form bi where [latex]i=\sqrt{-1}\\[/latex]

1. Explain how to add complex numbers.

2. What is the basic principle in multiplication of complex numbers?

3. Give an example to show the product of two imaginary numbers is not always imaginary.

4. What is a characteristic of the plot of a real number in the complex plane?

For the following exercises, evaluate the algebraic expressions.

5. [latex]\text{If }f\left(x\right)={x}^{2}+x - 4\\[/latex], evaluate [latex]f\left(2i\right)\\[/latex].

6. [latex]\text{If }f\left(x\right)={x}^{3}-2\\[/latex], evaluate [latex]f\left(i\right)\\[/latex].

7. [latex]\text{If }f\left(x\right)={x}^{2}+3x+5\\[/latex], evaluate [latex]f\left(2+i\right)\\[/latex].

8. [latex]\text{If }f\left(x\right)=2{x}^{2}+x - 3\\[/latex], evaluate [latex]f\left(2 - 3i\right)\\[/latex].

9. [latex]\text{If }f\left(x\right)=\frac{x+1}{2-x}\\[/latex], evaluate [latex]f\left(5i\right)\\[/latex].

10. [latex]\text{If }f\left(x\right)=\frac{1+2x}{x+3}\\[/latex], evaluate [latex]f\left(4i\right)\\[/latex].

For the following exercises, determine the number of real and nonreal solutions for each quadratic function shown.

11.

12.

For the following exercises, plot the complex numbers on the complex plane.

13. [latex]1 - 2i\\[/latex]

14. [latex]-2+3i\\[/latex]

15. i

16. [latex]-3 - 4i\\[/latex]

For the following exercises, perform the indicated operation and express the result as a simplified complex number.

17. [latex]\left(3+2i\right)+\left(5 - 3i\right)\\[/latex]

18. [latex]\left(-2 - 4i\right)+\left(1+6i\right)\\[/latex]

19. [latex]\left(-5+3i\right)-\left(6-i\right)\\[/latex]

20. [latex]\left(2 - 3i\right)-\left(3+2i\right)\\[/latex]

21. [latex]\left(-4+4i\right)-\left(-6+9i\right)\\[/latex]

22. [latex]\left(2+3i\right)\left(4i\right)\\[/latex]

23. [latex]\left(5 - 2i\right)\left(3i\right)\\[/latex]

24. [latex]\left(6 - 2i\right)\left(5\right)\\[/latex]

25. [latex]\left(-2+4i\right)\left(8\right)\\[/latex]

26. [latex]\left(2+3i\right)\left(4-i\right)\\[/latex]

27. [latex]\left(-1+2i\right)\left(-2+3i\right)\\[/latex]

28. [latex]\left(4 - 2i\right)\left(4+2i\right)\\[/latex]

29. [latex]\left(3+4i\right)\left(3 - 4i\right)\\[/latex]

30. [latex]\frac{3+4i}{2}\\[/latex]

31. [latex]\frac{6 - 2i}{3}\\[/latex]

32. [latex]\frac{-5+3i}{2i}\\[/latex]

33. [latex]\frac{6+4i}{i}\\[/latex]

34. [latex]\frac{2 - 3i}{4+3i}\\[/latex]

35. [latex]\frac{3+4i}{2-i}\\[/latex]

36. [latex]\frac{2+3i}{2 - 3i}\\[/latex]

37. [latex]\sqrt{-9}+3\sqrt{-16}\\[/latex]

38. [latex]-\sqrt{-4}-4\sqrt{-25}\\[/latex]

39. [latex]\frac{2+\sqrt{-12}}{2}\\[/latex]

40. [latex]\frac{4+\sqrt{-20}}{2}\\[/latex]

41. [latex]{i}^{8}\\[/latex]

42. [latex]{i}^{15}\\[/latex]

43. [latex]{i}^{22}\\[/latex]

For the following exercises, use a calculator to help answer the questions.

44. Evaluate [latex]{\left(1+i\right)}^{k}\\[/latex] for [latex]k=\text{4, 8, and 12}\text{.}\\[/latex] Predict the value if [latex]k=16\\[/latex].

45. Evaluate [latex]{\left(1-i\right)}^{k}\\[/latex] for [latex]k=\text{2, 6, and 10}\text{.}\\[/latex] Predict the value if [latex]k=14\\[/latex].

46. Evaluate [latex]\left(1+i\right)^{k}-\left(1-i\right)^{k}\\[/latex] for [latex]k=\text{4, 8, and 12}\\[/latex]. Predict the value for [latex]k=16\\[/latex].

47. Show that a solution of [latex]{x}^{6}+1=0\\[/latex] is [latex]\frac{\sqrt{3}}{2}+\frac{1}{2}i\\[/latex].

48. Show that a solution of [latex]{x}^{8}-1=0\\[/latex] is [latex]\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\\[/latex].

For the following exercises, evaluate the expressions, writing the result as a simplified complex number.

49. [latex]\frac{1}{i}+\frac{4}{{i}^{3}}\\[/latex]

50. [latex]\frac{1}{{i}^{11}}-\frac{1}{{i}^{21}}\\[/latex]

51. [latex]{i}^{7}\left(1+{i}^{2}\right)\\[/latex]

52. [latex]{i}^{-3}+5{i}^{7}\\[/latex]

53. [latex]\frac{\left(2+i\right)\left(4 - 2i\right)}{\left(1+i\right)}\\[/latex]

54. [latex]\frac{\left(1+3i\right)\left(2 - 4i\right)}{\left(1+2i\right)}\\[/latex]

55. [latex]\frac{{\left(3+i\right)}^{2}}{{\left(1+2i\right)}^{2}}\\[/latex]

56. [latex]\frac{3+2i}{2+i}+\left(4+3i\right)\\[/latex]

57. [latex]\frac{4+i}{i}+\frac{3 - 4i}{1-i}\\[/latex]

58. [latex]\frac{3+2i}{1+2i}-\frac{2 - 3i}{3+i}\\[/latex]

Figure 1

The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.

An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as [latex]2{x}^{2}+3x - 1=0[/latex] and [latex]{x}^{2}-4=0[/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if [latex]a\cdot b=0[/latex], then [latex]a=0[/latex] or [latex]b=0[/latex], where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression [latex]\left(x - 2\right)\left(x+3\right)[/latex] by multiplying the two factors together.

The product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[/latex], where a, b, and c are real numbers, and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], is 1. We have one method of factoring quadratic equations in this form.

Solution

To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[/latex].

[latex]\begin{array}{l}1\cdot \left(-6\right)\hfill \\ \left(-6\right)\cdot 1\hfill \\ 2\cdot \left(-3\right)\hfill \\ 3\cdot \left(-2\right)\hfill \end{array}[/latex]

The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

[latex]\left(x - 2\right)\left(x+3\right)=0[/latex]

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

[latex]\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&=0\hfill \\ \left(x - 2\right)\hfill&=0\hfill \\ x\hfill&=2\hfill \\ \left(x+3\right)\hfill&=0\hfill \\ x\hfill&=-3\hfill \end{array}[/latex]

The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex]. We can see how the solutions relate to the graph in Figure 2. The solutions are the x-intercepts of [latex]{x}^{2}+x - 6=0[/latex].

Figure 2

Factoring and Solving a Quadratic Equation of Higher Order

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

1. With the quadratic in standard form, [latex]a{x}^{2}+bx+c=0[/latex], multiply [latex]a\cdot c[/latex].
2. Find two numbers whose product equals [latex]ac[/latex] and whose sum equals [latex]b[/latex].
3. Rewrite the equation replacing the [latex]bx[/latex] term with two terms using the numbers found in step 1 as coefficients of x.
4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
5. Factor out the expression in parentheses.
6. Set the expressions equal to zero and solve for the variable.

Solution

First, multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

[latex]\begin{array}{l}1\cdot 36\hfill \\ 2\cdot 18\hfill \\ 3\cdot 12\hfill \\ 4\cdot 9\hfill \\ 6\cdot 6\hfill \end{array}[/latex]

The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the b term, [latex]15x[/latex], with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then factor the last two terms.

[latex]\begin{array}{l}4{x}^{2}+3x+12x+9=0\hfill \\ x\left(4x+3\right)+3\left(4x+3\right)=0\hfill \\ \left(4x+3\right)\left(x+3\right)=0\hfill \end{array}[/latex]

Solve using the zero-product property.

[latex]\begin{array}{l}\left(4x+3\right)\left(x+3\right)=0\hfill \\ \hfill \\ \left(4x+3\right)=0\hfill \\ x=-\frac{3}{4}\hfill \\ \hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}[/latex]

The solutions are [latex]x=-\frac{3}{4}[/latex], [latex]x=-3[/latex].

Figure 3

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.

1. Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right sign of the equal sign.
   [latex]{x}^{2}+4x=-1[/latex]
2. Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
   [latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
3. Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
   [latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
4. The left side of the equation can now be factored as a perfect square.
   [latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
5. Use the square root property and solve.
   [latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
6. The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].

The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[/latex] and obtain a positive a. Given [latex]a{x}^{2}+bx+c=0[/latex], [latex]a\ne 0[/latex], we will complete the square as follows:

1. First, move the constant term to the right side of the equal sign:
   [latex]a{x}^{2}+bx=-c[/latex]
2. As we want the leading coefficient to equal 1, divide through by a:
   [latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
3. Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign:
   [latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
   [latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
5. Now, use the square root property, which gives
   [latex]\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
6. Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus,
   [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, [latex]{b}^{2}-4ac[/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation.

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], where [latex]a[/latex] and [latex]b[/latex] refer to the legs of a right triangle adjacent to the [latex]90^\circ [/latex] angle, and [latex]c[/latex] refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

where [latex]a[/latex] and [latex]b[/latex] refer to the legs of a right triangle adjacent to the [latex]{90}^{\\circ }[/latex] angle, and [latex]c[/latex] refers to the hypotenuse, as shown in Figure 4.

Figure 4

Example 12: Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle in Figure 5.

Figure 5

As we have measurements for side b and the hypotenuse, the missing side is a.

Key Concepts

• Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions.
• Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method.
• Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution.
• Completing the square is a method of solving quadratic equations when the equation cannot be factored.
• A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation.
• The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.
• The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation.

Glossary

completing the square a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square

discriminant the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots.

Pythagorean Theorem a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems

quadratic equation an equation containing a second-degree polynomial; can be solved using multiple methods

quadratic formula a formula that will solve all quadratic equations

square root property one of the methods used to solve a quadratic equation, in which the [latex]{x}^{2}[/latex] term is isolated so that the square root of both sides of the equation can be taken to solve for x

zero-product property the property that formally states that multiplication by zero is zero, so that each factor of a quadratic equation can be set equal to zero to solve equations

1. How do we recognize when an equation is quadratic?

2. When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form [latex]a{x}^{2}+bx+c=0[/latex] we may graph the equation [latex]y=a{x}^{2}+bx+c[/latex] and have no zeroes (x-intercepts).

3. When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

4. In the quadratic formula, what is the name of the expression under the radical sign [latex]{b}^{2}-4ac[/latex], and how does it determine the number of and nature of our solutions?

5. Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

For the following exercises, solve the quadratic equation by factoring.

6. [latex]{x}^{2}+4x - 21=0[/latex]

7. [latex]{x}^{2}-9x+18=0[/latex]

8. [latex]2{x}^{2}+9x - 5=0[/latex]

9. [latex]6{x}^{2}+17x+5=0[/latex]

10. [latex]4{x}^{2}-12x+8=0[/latex]

11. [latex]3{x}^{2}-75=0[/latex]

12. [latex]8{x}^{2}+6x - 9=0[/latex]

13. [latex]4{x}^{2}=9[/latex]

14. [latex]2{x}^{2}+14x=36[/latex]

15. [latex]5{x}^{2}=5x+30[/latex]

16. [latex]4{x}^{2}=5x[/latex]

17. [latex]7{x}^{2}+3x=0[/latex]

18. [latex]\frac{x}{3}-\frac{9}{x}=2[/latex]

For the following exercises, solve the quadratic equation by using the square root property.

19. [latex]{x}^{2}=36[/latex]

20. [latex]{x}^{2}=49[/latex]

21. [latex]{\left(x - 1\right)}^{2}=25[/latex]

22. [latex]{\left(x - 3\right)}^{2}=7[/latex]

23. [latex]{\left(2x+1\right)}^{2}=9[/latex]

24. [latex]{\left(x - 5\right)}^{2}=4[/latex]

For the following exercises, solve the quadratic equation by completing the square. Show each step.

25. [latex]{x}^{2}-9x - 22=0[/latex]

26. [latex]2{x}^{2}-8x - 5=0[/latex]

27. [latex]{x}^{2}-6x=13[/latex]

28. [latex]{x}^{2}+\frac{2}{3}x-\frac{1}{3}=0[/latex]

29. [latex]2+z=6{z}^{2}[/latex]

30. [latex]6{p}^{2}+7p - 20=0[/latex]

31. [latex]2{x}^{2}-3x - 1=0[/latex]

For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.

32. [latex]2{x}^{2}-6x+7=0[/latex]

33. [latex]{x}^{2}+4x+7=0[/latex]

34. [latex]3{x}^{2}+5x - 8=0[/latex]

35. [latex]9{x}^{2}-30x+25=0[/latex]

36. [latex]2{x}^{2}-3x - 7=0[/latex]

37. [latex]6{x}^{2}-x - 2=0[/latex]

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.

38. [latex]2{x}^{2}+5x+3=0[/latex]

39. [latex]{x}^{2}+x=4[/latex]

40. [latex]2{x}^{2}-8x - 5=0[/latex]

41. [latex]3{x}^{2}-5x+1=0[/latex]

42. [latex]{x}^{2}+4x+2=0[/latex]

43. [latex]4+\frac{1}{x}-\frac{1}{{x}^{2}}=0[/latex]

For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x-intercepts) by using 2^nd CALC 2:zero. Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.

44. [latex]{\text{Y}}_{1}=4{x}^{2}+3x - 2[/latex]

45. [latex]{\text{Y}}_{1}=-3{x}^{2}+8x - 1[/latex]

46. [latex]{\text{Y}}_{1}=0.5{x}^{2}+x - 7[/latex]

47. To solve the quadratic equation [latex]{x}^{2}+5x - 7=4[/latex], we can graph these two equations

and find the points of intersection. Recall 2^nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

48. To solve the quadratic equation [latex]0.3{x}^{2}+2x - 4=2[/latex], we can graph these two equations

and find the points of intersection. Recall 2^nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

49. Beginning with the general form of a quadratic equation, [latex]a{x}^{2}+bx+c=0[/latex], solve for x by using the completing the square method , thus deriving the quadratic formula.

50. Show that the sum of the two solutions to the quadratic equation is [latex]\frac{-b}{a}[/latex].

51. A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft². Solve the quadratic equation to find the length and width.

52. Abercrombie and Fitch stock had a price given as [latex]P=0.2{t}^{2}-5.6t+50.2[/latex], where [latex]t[/latex] is the time in months from 1999 to 2001. ( [latex]t=1[/latex] is January 1999). Find the two months in which the price of the stock was $30.

53. Suppose that an equation is given [latex]p=-2{x}^{2}+280x - 1000[/latex], where [latex]x[/latex] represents the number of items sold at an auction and [latex]p[/latex] is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2^nd CALC maximum. To obtain a good window for the curve, set [latex]x[/latex] [0,200] and [latex]y[/latex] [0,10000].

54. A formula for the normal systolic blood pressure for a man age [latex]A[/latex], measured in mmHg, is given as [latex]P=0.006{A}^{2}-0.02A+120[/latex]. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

55. The cost function for a certain company is [latex]C=60x+300[/latex] and the revenue is given by [latex]R=100x - 0.5{x}^{2}[/latex]. Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.

56. A falling object travels a distance given by the formula [latex]d=5t+16{t}^{2}[/latex] ft., where [latex]t[/latex] is measured in seconds. How long will it take for the object to traveled 74 ft.?

57. A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft². Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.

58. An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, [latex]P[/latex], who contracted the flu [latex]t[/latex] days after it broke out is given by the model [latex]P=-{t}^{2}+13t+130[/latex], where [latex]1\le t\le 6[/latex]. Find the day that 160 students had the flu. Recall that the restriction on [latex]t[/latex] is at most 6.

Solutions to Try Its

1. [latex]\left(x - 6\right)\left(x+1\right)=0;x=6,x=-1[/latex]

2. [latex]\left(x - 7\right)\left(x+3\right)=0[/latex], [latex]x=7[/latex], [latex]x=-3[/latex].

3. [latex]\left(x+5\right)\left(x - 5\right)=0[/latex], [latex]x=-5[/latex], [latex]x=5[/latex].

4. [latex]\left(3x+2\right)\left(4x+1\right)=0[/latex], [latex]x=-\frac{2}{3}[/latex], [latex]x=-\frac{1}{4}[/latex]

5. [latex]x=0,x=-10,x=-1[/latex]

6. [latex]x=4\pm \sqrt{5}[/latex]

7. [latex]x=3\pm \sqrt{22}[/latex]

8. [latex]x=-\frac{2}{3}[/latex], [latex]x=\frac{1}{3}[/latex]

9. [latex]5[/latex] units

Solutions to Odd-Numbered Exercises

1. It is a second-degree equation (the highest variable exponent is 2).

3. We want to take advantage of the zero property of multiplication in the fact that if [latex]a\cdot b=0[/latex] then it must follow that each factor separately offers a solution to the product being zero: [latex]a=0\text{ }or\text{ b}=0[/latex].

5. One, when no linear term is present (no x term), such as [latex]{x}^{2}=16[/latex]. Two, when the equation is already in the form [latex]{\left(ax+b\right)}^{2}=d[/latex].

7. [latex]x=6[/latex], [latex]x=3[/latex]

9. [latex]x=\frac{-5}{2}[/latex], [latex]x=\frac{-1}{3}[/latex]

11. [latex]x=5[/latex], [latex]x=-5[/latex]

13. [latex]x=\frac{-3}{2}[/latex], [latex]x=\frac{3}{2}[/latex]

15. [latex]x=-2[/latex]

17. [latex]x=0[/latex], [latex]x=\frac{-3}{7}[/latex]

19. [latex]x=-6[/latex], [latex]x=6[/latex]

21. [latex]x=6[/latex], [latex]x=-4[/latex]

23. [latex]x=1[/latex], [latex]x=-2[/latex]

25. [latex]x=-2[/latex], [latex]x=11[/latex]

27. [latex]x=3\pm \sqrt{22}[/latex]

29. [latex]z=\frac{2}{3}\\[/latex], [latex]z=-\frac{1}{2}[/latex]

31. [latex]x=\frac{3\pm \sqrt{17}}{4}[/latex]

33. Not real

35. One rational

37. Two real; rational

39. [latex]x=\frac{-1\pm \sqrt{17}}{2}[/latex]

41. [latex]x=\frac{5\pm \sqrt{13}}{6}[/latex]

43. [latex]x=\frac{-1\pm \sqrt{17}}{8}[/latex]

45. [latex]x\approx 0.131[/latex] and [latex]x\approx 2.535[/latex]

47. [latex]x\approx -6.7[/latex] and [latex]x\approx 1.7[/latex]

49. [latex]\begin{array}{l}a{x}^{2}+bx+c \hfill& =0\hfill \\ {x}^{2}+\frac{b}{a}x \hfill& =\frac{-c}{a}\hfill \\ {x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}} \hfill& =\frac{-c}{a}+\frac{b}{4{a}^{2}}\hfill \\ {\left(x+\frac{b}{2a}\right)}^{2}\hfill& =\frac{{b}^{2}-4ac}{4{a}^{2}}\hfill \\ x+\frac{b}{2a}\hfill& =\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x\hfill& =\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]

51. [latex]x\left(x+10\right)=119[/latex]; 7 ft. and 17 ft.

53. maximum at [latex]x=70[/latex]

55. The quadratic equation would be [latex]\left(100x - 0.5{x}^{2}\right)-\left(60x+300\right)=300[/latex]. The two values of [latex]x[/latex] are 20 and 60.

57. 3 feet

We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\frac{1}{2}}[/latex] is another way of writing [latex]\sqrt{16}[/latex]; [latex]{8}^{\frac{1}{3}}[/latex] is another way of writing [latex]\text{ }\sqrt[3]{8}[/latex]. The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.

We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\frac{2}{3}\left(\frac{3}{2}\right)=1[/latex], [latex]3\left(\frac{1}{3}\right)=1[/latex], and so on.

We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

Analysis of the Solution

We can see the solutions on the graph in Figure 1. The x-coordinates of the points where the graph crosses the x-axis are the solutions–the x-intercepts. Notice on the graph that at the solution [latex]x=0[/latex], the graph touches the x-axis and bounces back. It does not cross the x-axis. This is typical of double solutions.

Figure 1

Solution

This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.

[latex]\begin{array}{l}{x}^{3}+{x}^{2}-9x - 9\hfill&=0\hfill \\ {x}^{2}\left(x+1\right)-9\left(x+1\right)\hfill&=0\hfill \\ \left({x}^{2}-9\right)\left(x+1\right)\hfill&=0\hfill \end{array}[/latex]

The grouping process ends here, as we can factor [latex]{x}^{2}-9[/latex] using the difference of squares formula.

[latex]\begin{array}{l}\left({x}^{2}-9\right)\left(x+1\right)\hfill&=0\hfill \\ \left(x - 3\right)\left(x+3\right)\left(x+1\right)\hfill&=0\hfill \\ x\hfill&=3\hfill \\ x\hfill&=-3\hfill \\ x\hfill&=-1\hfill \end{array}[/latex]

The solutions are [latex]x=3[/latex], [latex]x=-3[/latex], and [latex]x=-1[/latex]. Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in Figure 2.

Figure 2

Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as

Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.

Next, we will learn how to solve an absolute value equation. To solve an equation such as [latex]|2x - 6|=8[/latex], we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is [latex]8[/latex] or [latex]-8[/latex]. This leads to two different equations we can solve independently.

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic.

Solving Equations in Quadratic Form

Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include [latex]{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0[/latex], and [latex]{x}^{\frac{2}{3}}+4{x}^{\frac{1}{3}}+2=0[/latex]. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.

Solving Rational Equations Resulting in a Quadratic

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

Key Concepts

• Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1.
• Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping.
• We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index.
• To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value.
• Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve.
• Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form.

Glossary

absolute value equation an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression

equations in quadratic form equations with a power other than 2 but with a middle term with an exponent that is one-half the exponent of the leading term

extraneous solutions any solutions obtained that are not valid in the original equation

polynomial equation an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents

radical equation an equation containing at least one radical term where the variable is part of the radicand

1. In a radical equation, what does it mean if a number is an extraneous solution?

2. Explain why possible solutions must be checked in radical equations.

3. Your friend tries to calculate the value [latex]-{9}^{\frac{3}{2}}[/latex] and keeps getting an ERROR message. What mistake is he or she probably making?

4. Explain why [latex]|2x+5|=-7[/latex] has no solutions.

5. Explain how to change a rational exponent into the correct radical expression.

For the following exercises, solve the rational exponent equation. Use factoring where necessary.

6. [latex]{x}^{\frac{2}{3}}=16[/latex]

7. [latex]{x}^{\frac{3}{4}}=27[/latex]

8. [latex]2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0[/latex]

9. [latex]{\left(x - 1\right)}^{\frac{3}{4}}=8[/latex]

10. [latex]{\left(x+1\right)}^{\frac{2}{3}}=4[/latex]

11. [latex]{x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0[/latex]

12. [latex]{x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0[/latex]

For the following exercises, solve the following polynomial equations by grouping and factoring.

13. [latex]{x}^{3}+2{x}^{2}-x - 2=0[/latex]

14. [latex]3{x}^{3}-6{x}^{2}-27x+54=0[/latex]

15. [latex]4{y}^{3}-9y=0[/latex]

16. [latex]{x}^{3}+3{x}^{2}-25x - 75=0[/latex]

17. [latex]{m}^{3}+{m}^{2}-m - 1=0[/latex]

18. [latex]2{x}^{5}-14{x}^{3}=0[/latex]

19. [latex]5{x}^{3}+45x=2{x}^{2}+18[/latex]

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

20. [latex]\sqrt{3x - 1}-2=0[/latex]

21. [latex]\sqrt{x - 7}=5[/latex]

22. [latex]\sqrt{x - 1}=x - 7[/latex]

23. [latex]\sqrt{3t+5}=7[/latex]

24. [latex]\sqrt{t+1}+9=7[/latex]

25. [latex]\sqrt{12-x}=x[/latex]

26. [latex]\sqrt{2x+3}-\sqrt{x+2}=2[/latex]

27. [latex]\sqrt{3x+7}+\sqrt{x+2}=1[/latex]

28. [latex]\sqrt{2x+3}-\sqrt{x+1}=1[/latex]

For the following exercises, solve the equation involving absolute value.

29. [latex]|3x - 4|=8[/latex]

30. [latex]|2x - 3|=-2[/latex]

31. [latex]|1 - 4x|-1=5[/latex]

32. [latex]|4x+1|-3=6[/latex]

33. [latex]|2x - 1|-7=-2[/latex]

34. [latex]|2x+1|-2=-3[/latex]

35. [latex]|x+5|=0[/latex]

36. [latex]-|2x+1|=-3[/latex]

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

37. [latex]{x}^{4}-10{x}^{2}+9=0[/latex]

38. [latex]4{\left(t - 1\right)}^{2}-9\left(t - 1\right)=-2[/latex]

39. [latex]{\left({x}^{2}-1\right)}^{2}+\left({x}^{2}-1\right)-12=0[/latex]

40. [latex]{\left(x+1\right)}^{2}-8\left(x+1\right)-9=0[/latex]

41. [latex]{\left(x - 3\right)}^{2}-4=0[/latex]

For the following exercises, solve for the unknown variable.

42. [latex]{x}^{-2}-{x}^{-1}-12=0[/latex]

43. [latex]\sqrt{{|x|}^{2}}=x[/latex]

44. [latex]{t}^{25}-{t}^{5}+1=0[/latex]

45. [latex]|{x}^{2}+2x - 36|=12[/latex]

For the following exercises, use the model for the period of a pendulum, [latex]T[/latex], such that [latex]T=2\pi \sqrt{\frac{L}{g}}[/latex], where the length of the pendulum is L and the acceleration due to gravity is [latex]g[/latex].

46. If the acceleration due to gravity is [latex]9.8\mathrm{m/}{\text{s}}^{2}[/latex] and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).

47. If the gravity is [latex]32\frac{\text{ft}}{{\text{s}}^{2}}[/latex] and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.

For the following exercises, use a model for body surface area, BSA, such that [latex]BSA=\sqrt{\frac{wh}{3600}}[/latex], where w = weight in kg and h = height in cm.

48. Find the height of a 72-kg female to the nearest cm whose [latex]BSA=1.8[/latex].

49. Find the weight of a 177-cm male to the nearest kg whose [latex]BSA=2.1[/latex].

Solutions to Try Its

1. [latex]\frac{1}{4}[/latex]

2. [latex]25[/latex]

3. [latex]\{-1\}[/latex]

4. [latex]x=0[/latex], [latex]x=\frac{1}{2}[/latex], [latex]x=-\frac{1}{2}[/latex]

5. [latex]x=1[/latex]; extraneous solution [latex]x=-\frac{2}{9}[/latex]

6. [latex]x=-2[/latex]; extraneous solution [latex]x=-1[/latex]

7. [latex]x=-1[/latex], [latex]x=\frac{3}{2}[/latex]

8. [latex]x=-3,3,-i,i[/latex]

9. [latex]x=2,x=12[/latex]

10. [latex]x=-1[/latex], [latex]x=0[/latex] is not a solution.

Solutions to Odd-Numbered Exercises

1. This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

3. He or she is probably trying to enter negative 9, but taking the square root of [latex]-9[/latex] is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in [latex]-27[/latex].

5. A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

7. [latex]x=81[/latex]

9. [latex]x=17[/latex]

11. [latex]x=8, x=27[/latex]

13. [latex]x=-2,1,-1[/latex]

15. [latex]y=0, \frac{3}{2}, \frac{-3}{2}[/latex]

17. [latex]m=1,-1[/latex]

19. [latex]x=\frac{2}{5}[/latex]

21. [latex]x=32[/latex]

23. [latex]t=\frac{44}{3}[/latex]

25. [latex]x=3[/latex]

27. [latex]x=-2[/latex]

29. [latex]x=4,\frac{-4}{3}[/latex]

31. [latex]x=\frac{-5}{4},\frac{7}{4}[/latex]

33. [latex]x=3,-2[/latex]

35. [latex]x=-5[/latex]

37. [latex]x=1,-1,3,-3[/latex]

39. [latex]x=2,-2[/latex]

41. [latex]x=1,5[/latex]

43. All real numbers

45. [latex]x=4,6,-6,-8[/latex]

47. 10 in.

49. 90 kg

Figure 1

It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.

Indicating the solution to an inequality such as [latex]x\ge 4[/latex] can be achieved in several ways.

We can use a number line as shown in Figure 2. The blue ray begins at [latex]x=4[/latex] and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.

Figure 2

We can use set-builder notation: [latex]\{x|x\ge 4\}[/latex], which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.

The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to [latex]x\ge 4[/latex] are represented as [latex]\left[4,\infty \right)[/latex]. This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are [latex]\left[-2,6\right)[/latex], or all numbers between [latex]-2[/latex] and [latex]6[/latex], including [latex]-2[/latex], but not including [latex]6[/latex]; [latex]\left(-1,0\right)[/latex], all real numbers between, but not including [latex]-1[/latex] and [latex]0[/latex]; and [latex]\left(-\infty ,1\right][/latex], all real numbers less than and including [latex]1[/latex]. The table below outlines the possibilities.

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

A compound inequality includes two inequalities in one statement. A statement such as [latex]4<x\le 6[/latex] means [latex]4<x[/latex] and [latex]x\le 6[/latex]. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

Solution

Lets try the first method. Write two inequalities:

[latex]\begin{array}{lll}3+x> 7x - 2\hfill & \text{and}\hfill & 7x - 2> 5x - 10\hfill \\ 3> 6x - 2\hfill & \hfill & 2x - 2> -10\hfill \\ 5> 6x\hfill & \hfill & 2x> -8\hfill \\ \frac{5}{6}> x\hfill & \hfill & x> -4\hfill \\ x< \frac{5}{6}\hfill & \hfill & -4< x\hfill \end{array}[/latex]

The solution set is [latex]-4<x<\frac{5}{6}[/latex] or in interval notation [latex]\left(-4,\frac{5}{6}\right)[/latex]. Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line.

Figure 3

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at [latex]\left(-x,0\right)[/latex] has an absolute value of [latex]x[/latex], as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all [latex]x[/latex] –values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between [latex]x[/latex] and 600 is less than 200. We represent the distance between [latex]x[/latex] and 600 as [latex]|x - 600|[/latex], and therefore, [latex]|x - 600|\le 200[/latex] or

This means our returns would be between $400 and $800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

Solution

We want the distance between [latex]x[/latex] and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 4, to represent the condition to be satisfied.

Figure 4

The distance from [latex]x[/latex] to 5 can be represented using an absolute value symbol, [latex]|x - 5|[/latex]. Write the values of [latex]x[/latex] that satisfy the condition as an absolute value inequality.

[latex]|x - 5|\le 4[/latex]

We need to write two inequalities as there are always two solutions to an absolute value equation.

[latex]\begin{array}{lll}x - 5\le 4\hfill & \text{and}\hfill & x - 5\ge -4\hfill \\ x\le 9\hfill & \hfill & x\ge 1\hfill \end{array}[/latex]

If the solution set is [latex]x\le 9[/latex] and [latex]x\ge 1[/latex], then the solution set is an interval including all real numbers between and including 1 and 9.

So [latex]|x - 5|\le 4[/latex] is equivalent to [latex]\left[1,9\right][/latex] in interval notation.

Solution

We are trying to determine where [latex]y<0[/latex], which is when [latex]-\frac{1}{2}|4x - 5|+3<0[/latex]. We begin by isolating the absolute value.

[latex]\begin{array}{ll}-\frac{1}{2}|4x - 5|< -3\hfill & \text{Multiply both sides by -2, and reverse the inequality}.\hfill \\ |4x - 5|> 6\hfill & \hfill \end{array}[/latex]

Next, we solve for the equality [latex]|4x - 5|=6[/latex].

[latex]\begin{array}{lll}4x - 5=6\hfill & \hfill & 4x - 5=-6\hfill \\ 4x=11\hfill & \text{or}\hfill & 4x=-1\hfill \\ x=\frac{11}{4}\hfill & \hfill & x=-\frac{1}{4}\hfill \end{array}[/latex]

Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at [latex]x=-\frac{1}{4}[/latex] and [latex]x=\frac{11}{4}[/latex], and that the graph opens downward.

Figure 5

Key Concepts

• Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well.
• Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality.
• Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities.
• Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value.
• Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution.

Glossary

compound inequality a problem or a statement that includes two inequalities

interval an interval describes a set of numbers within which a solution falls

interval notation a mathematical statement that describes a solution set and uses parentheses or brackets to indicate where an interval begins and ends

linear inequality similar to a linear equation except that the solutions will include sets of numbers

1. When solving an inequality, explain what happened from Step 1 to Step 2:

2. When solving an inequality, we arrive at

Explain what our solution set is.

3. When writing our solution in interval notation, how do we represent all the real numbers?

4. When solving an inequality, we arrive at

Explain what our solution set is.

5. Describe how to graph [latex]y=|x - 3|[/latex]

For the following exercises, solve the inequality. Write your final answer in interval notation.

6. [latex]4x - 7\le 9[/latex]

7. [latex]3x+2\ge 7x - 1[/latex]

8. [latex]-2x+3>x - 5[/latex]

9. [latex]4\left(x+3\right)\ge 2x - 1[/latex]

10. [latex]-\frac{1}{2}x\le \frac{-5}{4}+\frac{2}{5}x[/latex]

11. [latex]-5\left(x - 1\right)+3>3x - 4-4x[/latex]

12. [latex]-3\left(2x+1\right)>-2\left(x+4\right)[/latex]

13. [latex]\frac{x+3}{8}-\frac{x+5}{5}\ge \frac{3}{10}[/latex]

14. [latex]\frac{x - 1}{3}+\frac{x+2}{5}\le \frac{3}{5}[/latex]

For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation.

15. [latex]|x+9|\ge -6[/latex]

16. [latex]|2x+3|<7[/latex]

17. [latex]|3x - 1|>11[/latex]

18. [latex]|2x+1|+1\le 6[/latex]

19. [latex]|x - 2|+4\ge 10[/latex]

20. [latex]|-2x+7|\le 13[/latex]

21. [latex]|x - 7|<-4[/latex]

22. [latex]|x - 20|>-1[/latex]

23. [latex]|\frac{x - 3}{4}|<2[/latex]

For the following exercises, describe all the x-values within or including a distance of the given values.

24. Distance of 5 units from the number 7

25. Distance of 3 units from the number 9

26. Distance of 10 units from the number 4

27. Distance of 11 units from the number 1

For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation.

28. [latex]-4<3x+2\le 18[/latex]

29. [latex]3x+1>2x - 5>x - 7[/latex]

30. [latex]3y<5 - 2y<7+y[/latex]

31. [latex]2x - 5<-11\text{ or }5x+1\ge 6[/latex]

32. [latex]x+7<x+2[/latex]

For the following exercises, graph the function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.

33. [latex]|x - 1|>2[/latex]

34. [latex]|x+3|\ge 5[/latex]

35. [latex]|x+7|\le 4[/latex]

36. [latex]|x - 2|<7[/latex]

37. [latex]|x - 2|<0[/latex]

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines.

38. [latex]x+3<3x - 4[/latex]

39. [latex]x - 2>2x+1[/latex]

40. [latex]x+1>x+4[/latex]

41. [latex]\frac{1}{2}x+1>\frac{1}{2}x - 5[/latex]

42. [latex]4x+1<\frac{1}{2}x+3[/latex]

For the following exercises, write the set in interval notation.

43. [latex]\{x|-1<x<3\}[/latex]

44. [latex]\{x|x\ge 7\}[/latex]

45. [latex]\{x|x<4\}[/latex]

46. [latex]\{x|x\text{ is all real numbers}\}[/latex]

For the following exercises, write the interval in set-builder notation.

47. [latex]\left(-\infty ,6\right)[/latex]

48. [latex]\left(4,+\infty \right)[/latex]

49. [latex]\left[-3,5\right)[/latex]

50. [latex]\left[-4,1\right]\cup \left[9,\infty \right)[/latex]

For the following exercises, write the set of numbers represented on the number line in interval notation.

51.

52.

53.

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter y2 = the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, 1:abs(. Find the points of intersection, recall (2^nd CALC 5:intersection, 1^st curve, enter, 2^nd curve, enter, guess, enter). Copy a sketch of the graph and shade the x-axis for your solution set to the inequality. Write final answers in interval notation.

54. [latex]|x+2|-5< 2[/latex]

55. [latex]\frac{-1}{2}|x+2|< 4[/latex]

56. [latex]|4x+1|-3> 2[/latex]

57. [latex]|x - 4|< 3[/latex]

58. [latex]|x+2|\ge 5[/latex]

59. Solve [latex]|3x+1|=|2x+3|[/latex]

60. Solve [latex]{x}^{2}-x>12[/latex]

61. [latex]\frac{x - 5}{x+7}\le 0[/latex], [latex]x\ne -7[/latex]

62. [latex]p=-{x}^{2}+130x - 3000[/latex] is a profit formula for a small business. Find the set of x-values that will keep this profit positive.

63. In chemistry the volume for a certain gas is given by [latex]V=20T[/latex], where V is measured in cc and T is temperature in ºC. If the temperature varies between 80ºC and 120ºC, find the set of volume values.

64. A basic cellular package costs $20/mo. for 60 min of calling, with an additional charge of $.30/min beyond that time. The cost formula would be [latex]C=\$20+.30\left(x - 60\right)[/latex]. If you have to keep your bill lower than $50, what is the maximum calling minutes you can use?

Solutions to Try Its

1. [latex]\left[-3,5\right][/latex]

2. [latex]\left(-\infty ,-2\right)\cup \left[3,\infty \right)[/latex]

3. [latex]x<1[/latex]

4. [latex]x\ge -5[/latex]

5. [latex]\left(2,\infty \right)[/latex]

6. [latex]\left[-\frac{3}{14},\infty \right)[/latex]

7. [latex]6<x\le 9\text{ }\text{ }\text{or}\left(6,9\right][/latex]

8. [latex]\left(-\frac{1}{8},\frac{1}{2}\right)[/latex]

9. [latex]|x - 2|\le 3[/latex]

10. [latex]k\le 1[/latex] or [latex]k\ge 7[/latex]; in interval notation, this would be [latex]\left(-\infty ,1\right]\cup \left[7,\infty \right)[/latex].

Solutions to Odd-Numbered Exercises

1. When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

3. [latex]\left(-\infty ,\infty \right)[/latex]

5. We start by finding the x-intercept, or where the function = 0. Once we have that point, which is [latex]\left(3,0\right)[/latex], we graph to the right the straight line graph [latex]y=x - 3[/latex], and then when we draw it to the left we plot positive y values, taking the absolute value of them.

7. [latex]\left(-\infty ,\frac{3}{4}\right][/latex]

9. [latex]\left[\frac{-13}{2},\infty \right)[/latex]

11. [latex]\left(-\infty ,3\right)[/latex]

13. [latex]\left(-\infty ,-\frac{37}{3}\right][/latex]

15. All real numbers [latex]\left(-\infty ,\infty \right)[/latex]

17. [latex]\left(-\infty ,\frac{-10}{3}\right)\cup \left(4,\infty \right)[/latex]

19. [latex]\left(-\infty ,-4\right]\cup \left[8,+\infty \right)[/latex]

21. No solution

23. [latex]\left(-5,11\right)[/latex]

25. [latex]\left[6,12\right][/latex]

27. [latex]\left[-10,12\right][/latex]

29. [latex]\begin{array}{ll}x> -6\text{ and }x> -2\hfill & \text{Take the intersection of two sets}.\hfill \\ x>-2,\text{ }\left(-2,+\infty \right)\hfill & \hfill \end{array}[/latex]

31. [latex]\begin{array}{ll}x< -3\text{ }\mathrm{or}\text{ }x\ge 1\hfill & \text{Take the union of the two sets}.\hfill \\ \left(-\infty ,-3\right){\cup }\left[1,\infty \right)\hfill & \hfill \end{array}[/latex]

33. [latex]\left(-\infty ,-1\right)\cup \left(3,\infty \right)[/latex]

35. [latex]\left[-11,-3\right][/latex]

37. It is never less than zero. No solution.

39. Where the blue line is above the orange line; point of intersection is [latex]x=-3[/latex].
[latex]\left(-\infty ,-3\right)[/latex]

41. Where the blue line is above the orange line; always. All real numbers.
[latex]\left(-\infty ,-\infty \right)[/latex]

43. [latex]\left(-1,3\right)[/latex]

45. [latex]\left(-\infty ,4\right)[/latex]

47. [latex]\{x|x<6\}[/latex]

49. [latex]\{x|-3\le x<5\}[/latex]

51. [latex]\left(-2,1\right][/latex]

53. [latex]\left(-\infty ,4\right][/latex]

55. Where the blue is below the orange; always. All real numbers. [latex]\left(-\infty ,+\infty \right)[/latex].

57. Where the blue is below the orange; [latex]\left(1,7\right)[/latex].

59. [latex]x=2,\frac{-4}{5}[/latex]

61. [latex]\left(-7,5\right][/latex]

63. [latex]\begin{array}{l}80\le T\le 120\\ 1,600\le 20T\le 2,400\end{array}[/latex]
[latex]\left[1,600, 2,400\right][/latex]

A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.

A relation is a set of ordered pairs. The set of the first components of each ordered pair is called the domain and the set of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.

The domain is [latex]\left\{1,2,3,4,5\right\}[/latex]. The range is [latex]\left\{2,4,6,8,10\right\}[/latex].

Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter [latex]x[/latex]. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter [latex]y[/latex].

A function [latex]f[/latex] is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, [latex]\left\{1,2,3,4,5\right\}[/latex], is paired with exactly one element in the range, [latex]\left\{2,4,6,8,10\right\}[/latex].

Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as

Notice that each element in the domain, [latex]\left\{\text{even,}\text{odd}\right\}[/latex] is not paired with exactly one element in the range, [latex]\left\{1,2,3,4,5\right\}[/latex]. For example, the term “odd” corresponds to three values from the domain, [latex]\left\{1,3,5\right\}[/latex] and the term “even” corresponds to two values from the range, [latex]\left\{2,4\right\}[/latex]. This violates the definition of a function, so this relation is not a function.

Figure 1 compares relations that are functions and not functions.

Figure 1. (a) This relationship is a function because each input is associated with a single output. Note that input [latex]q[/latex] and [latex]r[/latex] both give output [latex]n[/latex]. (b) This relationship is also a function. In this case, each input is associated with a single output. (c) This relationship is not a function because input [latex]q[/latex] is associated with two different outputs.

Example 1: Determining If Menu Price Lists Are Functions

The coffee shop menu, shown in Figure 2 consists of items and their prices.

1. Is price a function of the item?
2. Is the item a function of the price?

Figure 2

Watch this video online: https://youtu.be/zT69oxcMhPw

Using Function Notation

Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.

To represent “height is a function of age,” we start by identifying the descriptive variables [latex]h[/latex] for height and [latex]a[/latex] for age. The letters [latex]f,g[/latex], and [latex]h[/latex] are often used to represent functions just as we use [latex]x,y[/latex], and [latex]z[/latex] to represent numbers and [latex]A,B[/latex], and [latex]C[/latex] to represent sets.

[latex]\begin{cases}h\text{ is }f\text{ of }a\hfill & \hfill & \hfill & \hfill & \text{We name the function }f;\text{ height is a function of age}.\hfill \\ h=f\left(a\right)\hfill & \hfill & \hfill & \hfill & \text{We use parentheses to indicate the function input}\text{. }\hfill \\ f\left(a\right)\hfill & \hfill & \hfill & \hfill & \text{We name the function }f;\text{ the expression is read as ``}f\text{ of }a\text{.''}\hfill \end{cases}[/latex]

Remember, we can use any letter to name the function; the notation [latex]h\left(a\right)[/latex] shows us that [latex]h[/latex] depends on [latex]a[/latex]. The value [latex]a[/latex] must be put into the function [latex]h[/latex] to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.

We can also give an algebraic expression as the input to a function. For example [latex]f\left(a+b\right)[/latex] means “first add a and b, and the result is the input for the function f.” The operations must be performed in this order to obtain the correct result.

A General Note: Function Notation

The notation [latex]y=f\left(x\right)[/latex] defines a function named [latex]f[/latex]. This is read as [latex]``y[/latex] is a function of [latex]x.''[/latex] The letter [latex]x[/latex] represents the input value, or independent variable. The letter [latex]y\text{,\hspace{0.17em}}[/latex] or [latex]f\left(x\right)[/latex], represents the output value, or dependent variable.

Example 3: Using Function Notation for Days in a Month

Use function notation to represent a function whose input is the name of a month and output is the number of days in that month.

Solution

The number of days in a month is a function of the name of the month, so if we name the function [latex]f[/latex], we write [latex]\text{days}=f\left(\text{month}\right)[/latex] or [latex]d=f\left(m\right)[/latex]. The name of the month is the input to a “rule” that associates a specific number (the output) with each input.

Figure 4

For example, [latex]f\left(\text{March}\right)=31[/latex], because March has 31 days. The notation [latex]d=f\left(m\right)[/latex] reminds us that the number of days, [latex]d[/latex] (the output), is dependent on the name of the month, [latex]m[/latex] (the input).

Analysis of the Solution

Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs.

Example 4: Interpreting Function Notation

A function [latex]N=f\left(y\right)[/latex] gives the number of police officers, [latex]N[/latex], in a town in year [latex]y[/latex]. What does [latex]f\left(2005\right)=300[/latex] represent?

Solution

When we read [latex]f\left(2005\right)=300[/latex], we see that the input year is 2005. The value for the output, the number of police officers [latex]\left(N\right)[/latex], is 300. Remember, [latex]N=f\left(y\right)[/latex]. The statement [latex]f\left(2005\right)=300[/latex] tells us that in the year 2005 there were 300 police officers in the town.

Q & A

Instead of a notation such as [latex]y=f\left(x\right)[/latex], could we use the same symbol for the output as for the function, such as [latex]y=y\left(x\right)[/latex], meaning “y is a function of x?”

Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself we like to maintain a distinction between a function such as [latex]f[/latex], which is a rule or procedure, and the output [latex]y[/latex] we get by applying [latex]f[/latex] to a particular input [latex]x[/latex]. This is why we usually use notation such as [latex]y=f\left(x\right),P=W\left(d\right)[/latex], and so on.

Representing Functions Using Tables

A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship.

The table below lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function [latex]f[/latex] where [latex]D=f\left(m\right)[/latex] identifies months by an integer rather than by name.

Month number, [latex]m[/latex] (input)	1	2	3	4	5	6	7	8	9	10	11	12
Days in month, [latex]D[/latex] (output)	31	28	31	30	31	30	31	31	30	31	30	31

The table below defines a function [latex]Q=g\left(n\right)[/latex]. Remember, this notation tells us that [latex]g[/latex] is the name of the function that takes the input [latex]n[/latex] and gives the output [latex]Q\text{\hspace{0.17em}.}[/latex]

[latex]n[/latex]	1	2	3	4	5
[latex]Q[/latex]	8	6	7	6	8

The table below displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.

Age in years, [latex]\text{ }a\text{ }[/latex] (input)	5	5	6	7	8	9	10
Height in inches, [latex]\text{ }h\text{ }[/latex] (output)	40	42	44	47	50	52	54

How To: Given a table of input and output values, determine whether the table represents a function.

1. Identify the input and output values.
2. Check to see if each input value is paired with only one output value. If so, the table represents a function.

Example 5: Identifying Tables that Represent Functions

Which table, a), b), or c), represents a function (if any)?

a)

Table A

Input	Output
2	1
5	3
8	6

b)

Table B

Input	Output
–3	5
0	1
4	5

c)

Table C

Input	Output
1	0
5	2
5	4

Solution

a) and b) define functions. In both, each input value corresponds to exactly one output value. c) does not define a function because the input value of 5 corresponds to two different output values.

When a table represents a function, corresponding input and output values can also be specified using function notation.

The function represented by a) can be represented by writing

[latex]f\left(2\right)=1,f\left(5\right)=3,\text{and }f\left(8\right)=6[/latex]

Similarly, the statements [latex]g\left(-3\right)=5,g\left(0\right)=1,\text{and }g\left(4\right)=5[/latex] represent the function in b).

c) cannot be expressed in a similar way because it does not represent a function.

When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.

When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.

Evaluation of Functions in Algebraic Forms

 

When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function [latex]f\left(x\right)=5 - 3{x}^{2}[/latex] can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.

How To: Given the formula for a function, evaluate.

1. Replace the input variable in the formula with the value provided.
2. Calculate the result.

Example 6: Evaluating Functions

Given the function [latex]h\left(p\right)={p}^{2}+2p[/latex], evaluate [latex]h\left(4\right)[/latex].

Solution

To evaluate [latex]h\left(4\right)[/latex], we substitute the value 4 for the input variable [latex]p[/latex] in the given function.

[latex]\begin{cases}\text{ }h\left(p\right)={p}^{2}+2p\hfill \\ \text{ }h\left(4\right)={\left(4\right)}^{2}+2\left(4\right)\hfill \\ \text{ }=16+8\hfill \\ \text{ }=24\hfill \end{cases}[/latex]

Therefore, for an input of 4, we have an output of 24.

Watch this video online: https://youtu.be/Ehkzu5Uv7O0

Example 7: Evaluating Functions at Specific Values

Evaluate [latex]f\left(x\right)={x}^{2}+3x - 4[/latex] at

1. [latex]2[/latex]
2. [latex]a[/latex]
3. [latex]a+h[/latex]
4. [latex]\frac{f\left(a+h\right)-f\left(a\right)}{h}[/latex]

Solution

Replace the [latex]x[/latex] in the function with each specified value.

1. Because the input value is a number, 2, we can use algebra to simplify.
   [latex]\begin{cases}f\left(2\right)={2}^{2}+3\left(2\right)-4\hfill \\ =4+6 - 4\hfill \\ =6\hfill \end{cases}[/latex]
2. In this case, the input value is a letter so we cannot simplify the answer any further.
   [latex]f\left(a\right)={a}^{2}+3a - 4[/latex]
3. With an input value of [latex]a+h[/latex], we must use the distributive property.
   [latex]\begin{cases}f\left(a+h\right)={\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ ={a}^{2}+2ah+{h}^{2}+3a+3h - 4\hfill \end{cases}[/latex]
4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that
   [latex]f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h - 4[/latex]

   and we know that

   [latex]f\left(a\right)={a}^{2}+3a - 4[/latex]

   Now we combine the results and simplify.

   [latex]\begin{cases}\begin{cases}\hfill \\ \frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\right)-\left({a}^{2}+3a - 4\right)}{h}\hfill \end{cases}\hfill \\ \begin{cases}\text{ }=\frac{2ah+{h}^{2}+3h}{h}\hfill & \hfill \\ \text{ }=\frac{h\left(2a+h+3\right)}{h}\hfill & \begin{cases}{cc}\begin{cases}{cc}& \end{cases}& \end{cases}\text{Factor out }h.\hfill \\ \text{ }=2a+h+3\hfill & \begin{cases}{cc}\begin{cases}{cc}& \end{cases}& \end{cases}\text{Simplify}.\hfill \end{cases}\hfill \end{cases}[/latex]

Try It 2

Given the function [latex]g\left(m\right)=\sqrt{m - 4}[/latex], evaluate [latex]g\left(5\right)[/latex].

Solution

Watch this video online: https://youtu.be/GLOmTED1UwA

Example 8: Solving Functions

Given the function [latex]h\left(p\right)={p}^{2}+2p[/latex], solve for [latex]h\left(p\right)=3[/latex].

Solution

[latex]\begin{cases}\text{ }h\left(p\right)=3\hfill & \hfill & \hfill & \hfill \\ \text{ }{p}^{2}+2p=3\hfill & \hfill & \hfill & \text{Substitute the original function }h\left(p\right)={p}^{2}+2p.\hfill \\ \text{ }{p}^{2}+2p - 3=0\hfill & \hfill & \hfill & \text{Subtract 3 from each side}.\hfill \\ \text{ }\left(p+3\text{)(}p - 1\right)=0\hfill & \hfill & \hfill & \text{Factor}.\hfill \end{cases}[/latex]

If [latex]\left(p+3\right)\left(p - 1\right)=0[/latex], either [latex]\left(p+3\right)=0[/latex] or [latex]\left(p - 1\right)=0[/latex] (or both of them equal 0). We will set each factor equal to 0 and solve for [latex]p[/latex] in each case.

[latex]\begin{cases}\left(p+3\right)=0,\hfill & p=-3\hfill \\ \left(p - 1\right)=0,\hfill & p=1\hfill \end{cases}[/latex]

This gives us two solutions. The output [latex]h\left(p\right)=3[/latex] when the input is either [latex]p=1[/latex] or [latex]p=-3[/latex].

Figure 5

We can also verify by graphing as in Figure 5. The graph verifies that [latex]h\left(1\right)=h\left(-3\right)=3[/latex] and [latex]h\left(4\right)=24[/latex].

Watch this video online: https://youtu.be/NTmgEF_nZSc

Try It 3

Given the function [latex]g\left(m\right)=\sqrt{m - 4}[/latex], solve [latex]g\left(m\right)=2[/latex].

Solution

Evaluating Functions Expressed in Formulas

Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation [latex]2n+6p=12[/latex] expresses a functional relationship between [latex]n[/latex] and [latex]p[/latex]. We can rewrite it to decide if [latex]p[/latex] is a function of [latex]n[/latex].

How To: Given a function in equation form, write its algebraic formula.

1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable.
2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity.

Example 9: Finding an Equation of a Function

Express the relationship [latex]2n+6p=12[/latex] as a function [latex]p=f\left(n\right)[/latex], if possible.

Solution

To express the relationship in this form, we need to be able to write the relationship where [latex]p[/latex] is a function of [latex]n[/latex], which means writing it as p = expression involving n.
[latex]\begin{cases}2n+6p=12\hfill & \hfill \\ 6p=12 - 2n\hfill & \begin{cases}& & \end{cases}\text{Subtract }2n\text{ from both sides}.\hfill \\ p=\frac{12 - 2n}{6}\hfill & \begin{cases}& & \end{cases}\text{Divide both sides by 6 and simplify}.\hfill \\ p=\frac{12}{6}-\frac{2n}{6}\hfill & \hfill \\ p=2-\frac{1}{3}n\hfill & \hfill \end{cases}[/latex]

Therefore, [latex]p[/latex] as a function of [latex]n[/latex] is written as

[latex]p=f\left(n\right)=2-\frac{1}{3}n[/latex]

Watch this video online: https://youtu.be/lHTLjfPpFyQ

Analysis of the Solution

It is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula.

Example 10: Expressing the Equation of a Circle as a Function

Does the equation [latex]{x}^{2}+{y}^{2}=1[/latex] represent a function with [latex]x[/latex] as input and [latex]y[/latex] as output? If so, express the relationship as a function [latex]y=f\left(x\right)[/latex].

Solution

First we subtract [latex]{x}^{2}[/latex] from both sides.

[latex]{y}^{2}=1-{x}^{2}[/latex]

We now try to solve for [latex]y[/latex] in this equation.

[latex]\begin{cases}y=\pm \sqrt{1-{x}^{2}}\hfill \\ \text{ }=+\sqrt{1-{x}^{2}}\text{ and }-\sqrt{1-{x}^{2}}\hfill \end{cases}[/latex]

We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function

[latex]y=f\left(x\right)[/latex].

Try It 4

If [latex]x - 8{y}^{3}=0[/latex], express [latex]y[/latex] as a function of [latex]x[/latex].

Solution

Q & A

Are there relationships expressed by an equation that do represent a function but which still cannot be represented by an algebraic formula?

Yes, this can happen. For example, given the equation [latex]x=y+{2}^{y}[/latex], if we want to express [latex]y[/latex] as a function of [latex]x[/latex], there is no simple algebraic formula involving only [latex]x[/latex] that equals [latex]y[/latex]. However, each [latex]x[/latex] does determine a unique value for [latex]y[/latex], and there are mathematical procedures by which [latex]y[/latex] can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for [latex]y[/latex] as a function of [latex]x[/latex], even though the formula cannot be written explicitly.

Finding Function Values from a Graph

Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s).

Example 12: Reading Function Values from a Graph

Given the graph in Figure 6,

1. Evaluate [latex]f\left(2\right)[/latex].
2. Solve [latex]f\left(x\right)=4[/latex].

Figure 6

Solution

1. To evaluate [latex]f\left(2\right)[/latex], locate the point on the curve where [latex]x=2[/latex], then read the y-coordinate of that point. The point has coordinates [latex]\left(2,1\right)[/latex], so [latex]f\left(2\right)=1[/latex]. See Figure 7.
   
   
   

   

   Figure 7

2. To solve [latex]f\left(x\right)=4[/latex], we find the output value [latex]4[/latex] on the vertical axis. Moving horizontally along the line [latex]y=4[/latex], we locate two points of the curve with output value [latex]4:[/latex] [latex]\left(-1,4\right)[/latex] and [latex]\left(3,4\right)[/latex]. These points represent the two solutions to [latex]f\left(x\right)=4:[/latex] [latex]x=-1[/latex] or [latex]x=3[/latex]. This means [latex]f\left(-1\right)=4[/latex] and [latex]f\left(3\right)=4[/latex], or when the input is [latex]-1[/latex] or [latex]\text{3,}[/latex] the output is [latex]\text{4}\text{.}[/latex]See Figure 8.
   
   

   Figure 8

Try It 6

Using Figure 7, solve [latex]f\left(x\right)=1[/latex].

Solution

As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis.

The most common graphs name the input value [latex]x[/latex] and the output value [latex]y[/latex], and we say [latex]y[/latex] is a function of [latex]x[/latex], or [latex]y=f\left(x\right)[/latex] when the function is named [latex]f[/latex]. The graph of the function is the set of all points [latex]\left(x,y\right)[/latex] in the plane that satisfies the equation [latex]y=f\left(x\right)[/latex]. If the function is defined for only a few input values, then the graph of the function is only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 11 tell us that [latex]f\left(0\right)=2[/latex] and [latex]f\left(6\right)=1[/latex]. However, the set of all points [latex]\left(x,y\right)[/latex] satisfying [latex]y=f\left(x\right)[/latex] is a curve. The curve shown includes [latex]\left(0,2\right)[/latex] and [latex]\left(6,1\right)[/latex] because the curve passes through those points.

Figure 11

The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value.

Figure 12

Example 14: Applying the Vertical Line Test

Which of the graphs represent(s) a function [latex]y=f\left(x\right)?[/latex]

Figure 13

 

Solution

If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of Figure 13. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point.

Figure 14

Watch this video online: https://youtu.be/5Z8DaZPJLKY

Try It 8

Does the graph in Figure 15 represent a function?

Figure 15

Solution

Using the Horizontal Line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. How To: Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.

1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.
2. If there is any such line, determine that the function is not one-to-one.

Example 15: Applying the Horizontal Line Test

Consider the functions (a), and (b)shown in the graphs in Figure 16.

Figure 16

Are either of the functions one-to-one?

Solution

The function in (a) is not one-to-one. The horizontal line shown in Figure 17 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)

Figure 17

The function in (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.

Watch this video online: https://youtu.be/tbSGdcSN8RE

1. What is the difference between a relation and a function?

2. What is the difference between the input and the output of a function?

3. Why does the vertical line test tell us whether the graph of a relation represents a function?

4. How can you determine if a relation is a one-to-one function?

5. Why does the horizontal line test tell us whether the graph of a function is one-to-one?

For the following exercises, determine whether the relation represents a function.

6. [latex]\left\{\left(a,b\right),\text{ }\left(c,d\right),\text{ }\left(a,c\right)\right\}[/latex]

7. [latex]\left\{\left(a,b\right),\left(b,c\right),\left(c,c\right)\right\}[/latex]

For the following exercises, determine whether the relation represents [latex]y[/latex] as a function of [latex]x[/latex].

8. [latex]5x+2y=10[/latex]

9. [latex]y={x}^{2}[/latex]

10. [latex]x={y}^{2}[/latex]

11. [latex]3{x}^{2}+y=14[/latex]

12. [latex]2x+{y}^{2}=6[/latex]

13. [latex]y=-2{x}^{2}+40x[/latex]

14. [latex]y=\frac{1}{x}[/latex]

15. [latex]x=\frac{3y+5}{7y - 1}[/latex]

16. [latex]x=\sqrt{1-{y}^{2}}[/latex]

17. [latex]y=\frac{3x+5}{7x - 1}[/latex]

18. [latex]{x}^{2}+{y}^{2}=9[/latex]

19. [latex]2xy=1[/latex]

20. [latex]x={y}^{3}[/latex]

21. [latex]y={x}^{3}[/latex]

22. [latex]y=\sqrt{1-{x}^{2}}[/latex]

23. [latex]x=\pm \sqrt{1-y}[/latex]

24. [latex]y=\pm \sqrt{1-x}[/latex]

25. [latex]{y}^{2}={x}^{2}[/latex]

26. [latex]{y}^{3}={x}^{2}[/latex]

For the following exercises, evaluate the function [latex]f[/latex] at the indicated values [latex]\text{ }f\left(-3\right),f\left(2\right),f\left(-a\right),-f\left(a\right),f\left(a+h\right)[/latex].

27. [latex]f\left(x\right)=2x - 5[/latex]

28. [latex]f\left(x\right)=-5{x}^{2}+2x - 1[/latex]

29. [latex]f\left(x\right)=\sqrt{2-x}+5[/latex]

30. [latex]f\left(x\right)=\frac{6x - 1}{5x+2}[/latex]

31. [latex]f\left(x\right)=|x - 1|-|x+1|[/latex]

32. Given the function [latex]g\left(x\right)=5-{x}^{2}[/latex], evaluate [latex]\frac{g\left(x+h\right)-g\left(x\right)}{h},h\ne 0[/latex].

33. Given the function [latex]g\left(x\right)={x}^{2}+2x[/latex], evaluate [latex]\frac{g\left(x\right)-g\left(a\right)}{x-a},x\ne a[/latex].

34. Given the function [latex]k\left(t\right)=2t - 1:[/latex]

a. Evaluate [latex]k\left(2\right)[/latex].
b. Solve [latex]k\left(t\right)=7[/latex].

35. Given the function [latex]f\left(x\right)=8 - 3x:[/latex]

a. Evaluate [latex]f\left(-2\right)[/latex].
b. Solve [latex]f\left(x\right)=-1[/latex].

36. Given the function [latex]p\left(c\right)={c}^{2}+c:[/latex]

a. Evaluate [latex]p\left(-3\right)[/latex].
b. Solve [latex]p\left(c\right)=2[/latex].

37. Given the function [latex]f\left(x\right)={x}^{2}-3x:[/latex]

a. Evaluate [latex]f\left(5\right)[/latex].
b. Solve [latex]f\left(x\right)=4[/latex].

38. Given the function [latex]f\left(x\right)=\sqrt{x+2}:[/latex]

a. Evaluate [latex]f\left(7\right)[/latex].
b. Solve [latex]f\left(x\right)=4[/latex].

39. Consider the relationship [latex]3r+2t=18[/latex].

a. Write the relationship as a function [latex]r=f\left(t\right)[/latex].
b. Evaluate [latex]f\left(-3\right)[/latex].
c. Solve [latex]f\left(t\right)=2[/latex].

For the following exercises, use the vertical line test to determine which graphs show relations that are functions.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52. Given the following graph,

a. Evaluate [latex]f\left(-1\right)[/latex].
b. Solve for [latex]f\left(x\right)=3[/latex].

53. Given the following graph,

a. Evaluate [latex]f\left(0\right)[/latex].
b. Solve for [latex]f\left(x\right)=-3[/latex].

54. Given the following graph,

a. Evaluate [latex]f\left(4\right)[/latex].
b. Solve for [latex]f\left(x\right)=1[/latex].

For the following exercises, determine if the given graph is a one-to-one function.

55.

56.

57.

58.

59.

For the following exercises, determine whether the relation represents a function.

60. [latex]\left\{\left(-1,-1\right),\left(-2,-2\right),\left(-3,-3\right)\right\}[/latex]

61. [latex]\left\{\left(3,4\right),\left(4,5\right),\left(5,6\right)\right\}[/latex]

62. [latex]\left\{\left(2,5\right),\left(7,11\right),\left(15,8\right),\left(7,9\right)\right\}[/latex]

For the following exercises, determine if the relation represented in table form represents [latex]y[/latex] as a function of [latex]x[/latex].

63.

64.

Solutions to Try Its

1. a. yes; b. yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

2. [latex]g\left(5\right)=1[/latex]

3. [latex]m=8[/latex]

4. [latex]y=f\left(x\right)=\frac{\sqrt[3]{x}}{2}[/latex]

5. [latex]g\left(1\right)=8[/latex]

6. [latex]x=0[/latex] or [latex]x=2[/latex]

7. a. yes, because each bank account has a single balance at any given time; b. no, because several bank account numbers may have the same balance; c. no, because the same output may correspond to more than one input.

8. yes

Solutions to Odd-Numbered Exercises

1. A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

3. When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

5. When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

7. function

9. function

11. function

13. function

15. function

17. function

19. function

21. function

23. function

25. not a function

27. [latex]f\left(-3\right)=-11\\ f\left(2\right)=-1\\ f\left(-a\right)=-2a - 5\\-f\left(a\right)=-2a+5\\f\left(a+h\right)=2a+2h - 5\\[/latex]

29. [latex]f\left(-3\right)=\sqrt{5}+5\\ f\left(2\right)=5\\ f\left(-a\right)=\sqrt{2+a}+5\\ -f\left(a\right)=-\sqrt{2-a}-5\\f\left(a+h\right)=\\sqrt{2-a-h}+5\\[/latex]

31. [latex]f\left(-3\right)=2\\f\left(2\right)=1 - 3=-2\\f\left(-a\right)=|-a - 1|-|-a+1|\\ -f\left(a\right)=-|a - 1|+|a+1|\\\text{ }\text{ }f\left(a+h\right)=|a+h - 1|-|a+h+1|\\[/latex]

33. [latex]\frac{g\left(x\right)-g\left(a\right)}{x-a}=x+a+2,x\ne a[/latex]

35. a. [latex]f\left(-2\right)=14[/latex]; b. [latex]x=3[/latex]

37. a. [latex]f\left(5\right)=10[/latex]; b. [latex]x=-1\text{ }[/latex] or [latex]\text{ }x=4[/latex]

39. a. [latex]f\left(t\right)=6-\frac{2}{3}t[/latex]; b. [latex]f\left(-3\right)=8[/latex]; c. [latex]t=6[/latex]

41. not a function

43. function

45. function

47. function

49. function

51. function

53. a. [latex]f\left(0\right)=1[/latex]; b. [latex]f\left(x\right)=-3,x=-2\text{ }[/latex] or [latex]\text{ }x=2[/latex]

55. not a function so it is also not a one-to-one function

57. one-to-one function

59. function, but not one-to-one

61. function

63. function

65. not a function

67. [latex]f\left(x\right)=1,x=2[/latex]

69. [latex]f\left(-2\right)=14\\ f\left(-1\right)=11\\ f\left(0\right)=8\\f\left(1\right)=5\\ f\left(2\right)=2\\[/latex]

71. [latex]f\left(-2\right)=4\\\text{ }\\ f\left(-1\right)=4.414\\ f\left(0\right)=4.732\\ f\left(1\right)=4.5\\ f\left(2\right)=5.236\\[/latex]

73. [latex]f\left(-2\right)=\frac{1}{9}\\ f\left(-1\right)=\frac{1}{3}\\ f\left(0\right)=1\\ f\left(1\right)=3;\\f\left(2\right)=9\\[/latex]

75. 20

77. [latex]\left[0,\text{ 100}\right][/latex]

79.[latex]\left[-0.001,\text{ 0}\text{.001}\right][/latex]

81. [latex]\left[-1,000,000,\text{ 1,000,000}\right][/latex]

83. [latex]\left[0,\text{ 10}\right][/latex]

85. [latex]\left[-0.1,\text{0.1}\right][/latex]

87. [latex]\left[-100,\text{ 100}\right][/latex]

89. a. [latex]g\left(5000\right)=50[/latex]; b. The number of cubic yards of dirt required for a garden of 100 square feet is 1.

91. a. The height of a rocket above ground after 1 second is 200 ft. b. the height of a rocket above ground after 2 seconds is 350 ft.

If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these.

Figure 1. Based on data compiled by www.the-numbers.com.

In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.

Figure 2

We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products.

We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write [latex]\left(0,\text{ }100\right][/latex]. We will discuss interval notation in greater detail later.

Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative.

Before we begin, let us review the conventions of interval notation:

• The smallest term from the interval is written first.
• The largest term in the interval is written second, following a comma.
• Parentheses, ( or ), are used to signify that an endpoint is not included, called exclusive.
• Brackets, [ or ], are used to indicate that an endpoint is included, called inclusive.

The table below gives a summary of interval notation.

Solution

When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for [latex]x[/latex].

[latex]\begin{cases}2-x=0\hfill \\ -x=-2\hfill \\ x=2\hfill \end{cases}[/latex]

Now, we will exclude 2 from the domain. The answers are all real numbers where [latex]x<2[/latex] or [latex]x>2[/latex]. We can use a symbol known as the union, [latex]\cup [/latex], to combine the two sets. In interval notation, we write the solution: [latex]\left(\mathrm{-\infty },2\right)\cup \left(2,\infty \right)[/latex].

Figure 3

In interval form, the domain of [latex]f[/latex] is [latex]\left(-\infty ,2\right)\cup \left(2,\infty \right)[/latex].

Watch this video online: https://youtu.be/v0IhvIzCc_I

Watch this video online: https://youtu.be/lj_JB8sfyIM

In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, [latex]\left\{x|10\le x<30\right\}[/latex] describes the behavior of [latex]x[/latex] in set-builder notation. The braces { } are read as “the set of,” and the vertical bar | is read as “such that,” so we would read [latex]\left\{x|10\le x<30\right\}[/latex] as “the set of x-values such that 10 is less than or equal to [latex]x[/latex], and [latex]x[/latex] is less than 30.”

The table below compares inequality notation, set-builder notation, and interval notation.

	Inequality Notation	Set-builder Notation	Interval Notation
	5 < h ≤ 10	{ h | 5 < h ≤ 10}	(5, 10]
	5 ≤ h < 10	{ h | 5 ≤ h < 10}	[5, 10]
	5 < h < 10	{ h | 5 < 10 }	(5, 10)
	h < 10	{ h | h < 10 }	( −∞, 10)
	h ≥ 10	{ h | h ≥ 10 }	[10, ∞ )
	All real numbers	ℝ	( −∞, ∞ )

To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, [latex]\cup [/latex], to combine two unconnected intervals. For example, the union of the sets [latex]\left\{2,3,5\right\}[/latex] and [latex]\left\{4,6\right\}[/latex] is the set [latex]\left\{2,3,4,5,6\right\}[/latex]. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is

This video describes how to use interval notation to describe a set.

Watch this video online: https://youtu.be/hqg85P0ZMZ4

This video describes how to use Set-Builder notation to describe a set.

Watch this video online: https://youtu.be/rPcGeaDRnyc

Example 5: Describing Sets on the Real-Number Line

Describe the intervals of values shown in Figure 4 using inequality notation, set-builder notation, and interval notation.

Figure 4

Solution

To describe the values, [latex]x[/latex], included in the intervals shown, we would say, ” [latex]x[/latex] is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.”

Inequality	[latex]1\le x\le 3\text{or}x>5[/latex]
Set-builder notation	[latex]\left\{x|1\le x\le 3\text{or}x>5\right\}[/latex]
Interval notation	[latex]\left[1,3\right]\cup \left(5,\infty \right)[/latex]

Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set.

Try It 5

Given Figure 5, specify the graphed set in

1. words
2. set-builder notation
3. interval notation

Figure 5

Solution

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 6.

Figure 6

We can observe that the graph extends horizontally from [latex]-5[/latex] to the right without bound, so the domain is [latex]\left[-5,\infty \right)[/latex]. The vertical extent of the graph is all range values [latex]5[/latex] and below, so the range is [latex]\left(\mathrm{-\infty },5\right][/latex]. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.

Example 6: Finding Domain and Range from a Graph

Find the domain and range of the function [latex]f[/latex] whose graph is shown in Figure 7.

Figure 7

Solution

We can observe that the horizontal extent of the graph is –3 to 1, so the domain of [latex]f[/latex] is [latex]\left(-3,1\right][/latex].

Figure 8

The vertical extent of the graph is 0 to –4, so the range is [latex]\left[-4,0\right)[/latex].

Watch this video online: https://youtu.be/QAxZEelInJc

Example 7: Finding Domain and Range from a Graph of Oil Production

Find the domain and range of the function [latex]f[/latex] whose graph is shown in Figure 9.

Figure 9. (credit: modification of work by the U.S. Energy Information Administration)

Solution

The input quantity along the horizontal axis is “years,” which we represent with the variable [latex]t[/latex] for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable [latex]b[/latex] for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as [latex]1973\le t\le 2008[/latex] and the range as approximately [latex]180\le b\le 2010[/latex].

In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines.

Try It 6

Given the graph in Figure 10, identify the domain and range using interval notation.

Figure 10

Solution

Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function [latex]f\left(x\right)=|x|[/latex]. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0.

If we input 0, or a positive value, the output is the same as the input.

If we input a negative value, the output is the opposite of the input.

Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain.

We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income, S, would be 0.1S if [latex]{S}\le\[/latex] $10,000 and 1000 + 0.2 (S – $10,000), if S> $10,000.

Analysis of the Solution

The function is represented in Figure 21. The graph is a diagonal line from [latex]n=0[/latex] to [latex]n=10[/latex] and a constant after that. In this example, the two formulas agree at the meeting point where [latex]n=10[/latex], but not all piecewise functions have this property.

Figure 21

Watch this video online: https://youtu.be/B1jfpiI-QQ8

Example 12: Working with a Piecewise Function

A cell phone company uses the function below to determine the cost, [latex]C[/latex], in dollars for [latex]g[/latex] gigabytes of data transfer.

[latex]C\left(g\right)=\begin{cases}{25} \text{ if }{ 0 }<{ g }<{ 2 }\\ { 25+10 }\left(g - 2\right) \text{ if }{ g}\ge{ 2 }\end{cases}[/latex]

Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.

Solution

To find the cost of using 1.5 gigabytes of data, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula.

C(1.5) = $25

To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula.

C(4)=25 + 10( 4-2) =$45

Analysis of the Solution

The function is represented in Figure 22. We can see where the function changes from a constant to a shifted and stretched identity at [latex]g=2[/latex]. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain.

Figure 22

Example 13: Graphing a Piecewise Function

Sketch a graph of the function.

[latex]f\left(x\right)=\begin{cases}{ x }^{2} \text{ if }{ x }\le{ 1 }\\ { 3 } \text{ if } { 1 }&lt{ x }\le 2\\ { x } \text{ if }{ x }&gt{ 2 }\end{cases}[/latex]

Solution

Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality.

Below are the three components of the piecewise function graphed on separate coordinate systems.

(a) [latex]f\left(x\right)={x}^{2}\text{ if }x\le 1[/latex]; (b) [latex]f\left(x\right)=3\text{ if 1< }x\le 2[/latex]; (c) [latex]f\left(x\right)=x\text{ if }x>2[/latex]

Figure 23

Now that we have sketched each piece individually, we combine them in the same coordinate plane.

Figure 24

Analysis of the Solution

Note that the graph does pass the vertical line test even at [latex]x=1[/latex] and [latex]x=2[/latex] because the points [latex]\left(1,3\right)[/latex] and [latex]\left(2,2\right)[/latex] are not part of the graph of the function, though [latex]\left(1,1\right)[/latex] and [latex]\left(2,3\right)[/latex] are.

Key Concepts

• The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number.
• The domain of a function can be determined by listing the input values of a set of ordered pairs.
• The domain of a function can also be determined by identifying the input values of a function written as an equation.
• Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation.
• For many functions, the domain and range can be determined from a graph.
• An understanding of toolkit functions can be used to find the domain and range of related functions.
• A piecewise function is described by more than one formula.
• A piecewise function can be graphed using each algebraic formula on its assigned subdomain.

Glossary

interval notation

a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion

piecewise function

a function in which more than one formula is used to define the output

set-builder notation

a method of describing a set by a rule that all of its members obey; it takes the form [latex]\left\{x|\text{statement about }x\right\}[/latex]