Chapter Fifteen — Functions

The Humongous Book of Algebra Problems

326

Note: Problems 15.7–15.8 refer to the relation s(x) = 5 ± x.

15.8 Is s(x) a function? Why or why not?

According to Problem 15.7, s(–1) = 4 or s(–1) = 6. For a relation to be a

function, each value substituted into the relation must produce only one result.

In this case, substituting x = –1 into s(x) produces two results (4 and 6), so s(x)

is not a function.

Operations on Functions

+, –,

˙

, and ÷ functions

Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and

h(x) = 2x

2

– 13x + 20.

15.9 Express (f + g)(x) in terms of x.

The function (f + g)(x) represents the sum f(x) + g(x).

Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and

h(x) = 2x

2

– 13x + 20.

15.10 Evaluate (f + 4g – h)( –2).

The function (f + 4g – h)(x) represents the sum f(x) + 4g(x) – h(x). Evaluate

f(x), g(x), and h(x) for x = –2.

Use the values of f(–2), g(–2), and h(–2) to calculate (f + 4g – h)( –2).

This is

NOT the

distributive

property—you’re not

distributing x through

(f + g) to get f(x) + g(x).

It’s just notation:

(f + g)(x) = f(x) + g(x),

(f – g)(x) = f(x) – g(x),

(fg)(x) = f(x)

˙

g(x), and

.

Chapter Fifteen — Functions

The Humongous Book of Algebra Problems

327

Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and

h(x) = 2x

2

– 13x + 20.

15.11 Evaluate (fg)(6).

The function (fg)(x) is deﬁned as the product f(x) · g(x). Therefore,

(fg)(6) = f(6) · g(6). Evaluate f(x) and g(x) for x = 6.

Use the values of f(6) and g(6) to calculate (fg)(6).

Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and

h(x) = 2x

2

– 13x + 20.

15.12 Express (fg)(x) in terms of x and evaluate that expression for x = 6 to verify the

solution to Problem 15.11.

The function (fg)(x) is deﬁned as the product f(x) · g(x). Calculate the product.

Evaluate the expression for x = 6.

To calculate

(fg)(6), you can

either plug 6 into

f(x) and g(x) and then

multiply those results (like

Problem 15.11) or multiply

f(x) and g(x) rst and

THEN plug in x = 6 (like

Problem 15.12).

Chapter Fifteen — Functions

The Humongous Book of Algebra Problems

328

Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and

h(x) = 2x

2

– 13x + 20.

15.13 Express in terms of x.

The function is deﬁned as the quotient of h(x) and f(x).

Reduce the fraction to lowest terms by factoring the numerator.

Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and

h(x) = 2x

2

– 13x + 20.

15.14 Evaluate .

Substitute x = –1 into g(x) and h(x).

Calculate the quotient of h(–1) and g(–1).

Chapters

20 and 21

go into a lot

more detail about

rational expressions

(fractions that contain

polynomials), including

how to simplify

them.

The original

denominator of

the fraction was

x – 4, so when x = 4 the

denominator is 4 – 4 = 0.

Dividing by 0 is not

allowed. Sure you can

simplify the fraction to

get 2x – 5, but

is STILL undened

at x = 4.

Chapter Fifteen — Functions

The Humongous Book of Algebra Problems

329

Note: Problems 15.15–15.17 refer to the following functions: and

.

15.15 Evaluate (p + r)(4).

Calculate the sum p(4) + r(4).

Simplify the radical expressions.

Therefore, .

Note: Problems 15.15–15.17 refer to the following functions: and

.

15.16 Express (pr)(x) in terms of x and evaluate (pr)(–2).

Multiply functions p(x) and r(x).

Evaluate (pr)(–2) by substituting x = –2 into (pr)(x).

You can’t

add these yet

because they’re

not like radicals—

they don’t have

the same numbers

under the

radical symbol.

If you’re multiplying

two radicals with the

same index, you can

multiply the contents of

the radicals inside one

big radical sign with a

matching index.

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