Chapter 5 Inverse Functions and Applications

Section 5.2

Section 5.2 Composition of Functions

• Algebra of Functionso Evaluating functions that are sums, differences, products, and quotients algebraically and graphically.

• Composition of Functionso Finding compositions of functions algebraically and graphically.

• Applications

Algebra of Functions

Let f(x) and g(x) be two functions.

Sum f + g = (f + g)(x) = f(x) + g(x)

Difference f – g = (f – g)(x) = f(x) – g(x)

Product f g = (f g)(x) = f(x) g(x)

Quotient , where g(x) 0

The domain of the new functions sum, difference, product, and quotient of f and g will be the set of real numbers x contained in the domains of both f and g. For the quotient function we have the restriction that g(x) 0.

)x(g)x(f

xgf

gf

Let f(x) = 3x2 + x and g(x) = 2x – 6. Find each function and state its domain.

a. (f + g)(x) b. (f – g)(x) c. (f g)(x) d.

a. (f + g)(x) = f(x) + g(x) = (3x2 + x) + (2x – 6) = 3x2 + 3x – 6

The domain of the sum function is the set of real numbers xcontained in the domains of both f and g. Thus, the domainis (–∞, ∞).

b. (f – g)(x) = f(x) – g(x) = (3x2 + x) – (2x – 6) = 3x2 – x + 6

The domain of the difference function is the set of real numbers x contained in the domains of both f and g. Thus, the domainis (–∞, ∞).

(continued on the next slide)

xgf

(Contd.)

c. (f g)(x) = f(x) g(x) = (3x2 + x)(2x – 6) = 6x3 – 16x2 – 6x

The domain of the product function is the set of real numbers x contained in the domains of both f and g. Thus, the domain is (–∞, ∞).

d.

The domain of the quotient function will have the restriction that g(x) 0. Thus, the domain is the set of real numbers x such that x 3, or (–∞, 3) U (3, ∞).

6x2xx3

)x(g)x(f

xgf 2

Let f(x) = 2x – 1 and g(x) = 5x + 2. Evaluate (f + g)(3). We can follow any of these approaches:

Option 1: Determine (f + g)(x), then find (f + g)(3).

or

Option 2: Find f(3) and g(3), then add the resulting values.

Option 1:

(f + g)(x) = (2x – 1) + (5x + 2) = 7x + 1

Hence, (f + g)(3) = 7(3) + 1 = 22

Option 2:

f(3) = 2(3) – 1 = 5 and g(3) = 5(3) + 2 = 17

So, (f + g)(3) = 5 + 17 = 22

Let f(x) = 2x – 1 and g(x) = 5x + 2. Evaluate (f g)(0). We can follow any of these approaches:

Option 1: Determine (f g)(x), then find (f g)(0).

or

Option 2: Find f(0) and g(0), then multiply the resulting values.

Option 1:

(f g)(x) = (2x – 1)(5x + 2) = 10x2 – x – 2

(f g)(0) = 10(0)2 – (0) – 2 = –2

Option 2:

f(0) = 2(0) – 1 = –1 and g(0) = 5(0) + 2 = 2

So, (f g)(0) = (–1)(2) = –2

Use the given graph to evaluate each function:

a. (f + g)(9) b. (f – g)(0) c.

a.f(9) = –3 and g(9) = 8

(f + g)(9) = –3 + 8 = 5

b.f(0) = 0 and g(0) = –1

(f – g)(0) = 0 – (–1) = 1

c.f(1) = –1 and g(1) = 0

1gf

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1gf

Members of a martial arts club are planning a fundraising project to raise money to improve their facilities. The fundraising includes pizzas, sodas, and desert combo sales. The cost, C, of the project can be approximated by C(x) = 0.05x2 – 20x + 50, and the expected revenue generated for x combos sold is given by R(x) = 6.50x.

a. Find the profit function, P(x).

The profit is found by subtracting the cost from the revenue.

Therefore, the difference function R(x) – C(x) will give us thefundraising's profit.

R(x) – C(x) = 6.50x – (0.05x2 – 20x + 50)

= –0.05x2 + 26.5x – 50 = P(x)

(continued on the next slide)

(Contd.)

b. Find and interpret P(325).

We know P(x) = –0.05x2 + 26.5x – 50

P(325) = –0.05(325)2 + 26.5(325) – 50 = 3281.25

If the members of the martial arts club sell 325 food combos, they can expect a profit of $3,281.25.

Composition of Functions - Introduction

Composition of functions is the process of combining functions such that the output from one function becomes the input for the next function. This process is frequently referred to as "a function of a function.”

Suppose we have f(x) = 2x2 + 5 and g(x) = 3x.

The “composition of f with g” means that we will create a new function f(g(x)), where the input for f will be function g.

f(g(x)) = f(3x) = 2(3x)2 + 5 = 2(9x2) + 5 = 18x2 + 5

Observe that in f(g(x)), f is the outside function and g is the inside function.

Let f(x) = 2x2 + 5 and g(x) = 3x.

In the previous example, we did the “composition of f with g,” that is, f(g(x)).

Suppose we now want to determine the “composition of g with f.”

In this case, we will create a new function g(f(x)), where the input for g will be function f.

g(f(x)) = g(2x2 + 5) = 3(2x2 + 5) = 6x2 + 15

Notice that now we have g as the outside function, while f is the inside function.

Composition of Functions

Let f and g be two functions.

The composite function “f of g” is denoted f ◦ g and is defined as (f ◦ g)(x) = f(g(x)), provided that x is in the domain of g and g(x) is in the domain of f.

The composite function “g of f” is denoted g ◦ f and is defined as (g ◦ f)(x) = g(f(x)), provided that x is in the domain of f and f(x) is in the domain of g.

Composition of Functions – Some Notes

Do not confuse f ◦ g with the product of two functions, f g.

For most functions, composition is not commutative, that is, f ◦ g g ◦ f.

The order is important: 

(f ◦ g)(x) = f(g(x))   g function is inside and f function is outside

Evaluate f with g.

(g ◦ f)(x) = g(f(x)) f function is inside and g function is outside Evaluate g with f.  

Given f(x) = 3x – 1 and g(x) = x2 + 6, find each of the following. State the domain of each composite function.

a. (f ◦ g)(x) b. (g ◦ f)(x)

a. (f ◦ g)(x) = f(g(x))

Function g will be the input for function f.

f(g(x)) = f(x2 + 6) = 3(x2 + 6) – 1

= 3x2 + 18 – 1

= 3x2 + 17

There are no restrictions and the domain is (–∞, ∞).

(continued on the next slide)

 

 

 

(Contd.)

Given f(x) = 3x – 1 and g(x) = x2 + 6, find each of the following. State the domain of each composite function.

a. (f ◦ g)(x) b. (g ◦ f)(x)

b. (g ◦ f)(x) = g(f(x))

Function f will be the input for function g.

g(f(x)) = g(3x – 1) = (3x – 1)2 + 6

= (9x2 – 6x + 1) + 6

= 9x2 – 6x + 7

There are no restrictions and the domain is (–∞, ∞).

 

 

 

Given and , find f ◦ h.

State the domain of the composite function.

 

 

The domain is given by the interval (–6, ∞).

 

6x1

)x(h

x)x(f

6x1

f))x(h(f)x)(hf(

6x1

6x1

Let f(x) = 2x – 1 and g(x) = 5x + 2. Find (f ◦ g)(3). We can follow any of these approaches:

Option 1: Find the composite function in terms of x, then evaluate (f ◦ g)(x) when x = 3.

or

Option 2: Evaluate the inside function g when x = 3, then substitute the resulting value into the outside function f and evaluate. That is, the output for g(3) becomes the input for f.

Option 1:

(f ◦ g)(x) = f(g(x))

= f(5x + 2) = 2(5x + 2) – 1 = 10x + 4 – 1 = 10x + 3

Then, (f ◦ g)(3) = 10(3) + 3 = 33

(continued on the next slide)

(Contd.)

Let f(x) = 2x – 1 and g(x) = 5x + 2. Find (f ◦ g)(3). Option 2:

First, we evaluate g(3):

g(3) = 5(3) + 2 = 15 + 2 = 17

The output “17” becomes the input for f. Therefore, we now evaluate f(17):

f(17) = 2(17) – 1 = 34 – 1 = 33

Let f(x) = 7 – x, g(x) = x2 + 2x, and h(x) = x – 1. Find (f ◦ g ◦ h)(0.5). (f ◦ g ◦ h)(0.5) implies that we must first find (g ◦ h)(0.5).

To find (g ◦ h)(0.5), we start by evaluating h(0.5):

h(0.5) = (0.5) – 1 = –0.5

This output becomes the input for g:

g(–0.5 ) = (–0.5)2 + 2(–0.5)= –0.75

So, the result of the composition (g ◦ h)(0.5) = –0.75.

Finally, we find f(–0.75):

f(–0.75) = 7 – (–0.75) = 7.75

Hence, (f ◦ g ◦ h)(0.5) = 7.75

Use the given graph to find each composition:

a. (f ◦ g)(5) b. (g ◦ f)(0)

a.First, we have g(5) = 4.

Then we evaluate f(4)= –2

So, (f ◦ g)(5) = –2

b.(g ◦ f)(0) = g(f(0))

f(0) = 0

So, g(f(0)) = g(0)= –1

Pierre wants to buy a folding kayak for his summer vacations. The kayak he wants has a regular price of x dollars, but it is being sold with a rebate of $350 and a 25% discount.

a. Find a function, K(x), for the price of the kayak after the rebate.

K(x) = x – 350

b. Find a function, D(x), for the price of the kayak after the discount.

D(x) = 0.75x

c. Find (K ◦ D)(x).

(K ◦ D)(x) = K(D(x)) = K(0.75x) = 0.75x – 350

(continued on the next slide)

 

 

 

(Contd.)

d. Find and interpret (K ◦ D)(1595).

We know (K ◦ D)(x) = 0.75x – 350.

Thus, (K ◦ D)(1595) = 0.75(1595) – 350 = 846.25.

This is the kayak’s price after the discount followed by the rebate.

e. Find (D ◦ K)(x).

(D ◦ K)(x) = D(K(x))

We know D(x) = 0.75x and K(x) = x – 350.

So, D(K(x)) = D(x – 350) = 0.75(x – 300).

(continued on the next slide)

 

 

 

(Contd.)

f. Find and interpret (D ◦ K)(1595).

We know D(K(x)) = 0.75(x – 300).

Thus, (D ◦ K)(1595) = 0.75(1595 – 350) = 933.75.

This is the kayak’s price after the rebate followed by the discount.

g. Which composite function is the better option for Pierre,K ◦ D or D ◦ K? Explain your decision.

K ◦ D is the better option. Taking the 25% discount of the kayak's regular price first results in a greater price reduction.

 

 

 

Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 5.2.