Lesson 4-2 Operations on Functions Objective: To perform operations on functions and to determine the domain of the resulting functions.

Lesson 4-2 Operations on Functions

Objective: To perform operations on functions and to

determine the domain of the resulting functions.

Suppose a company manufactures and sells a certain product. If the cost of manufacturing x items of the product is given by the function C(x) and the revenue generated by the sale of the x items is given by the function R(x) then the company’s profit is given by the function P(x) where

P(x) = R(x) - C(x)That is the profit function is the difference between

the revenue and cost function. As the example from economics suggests it is possible to combine two given functions to produce a new function.

The Sum, Difference,Product and Quotient of FunctionsEach function listed below is defined for all x in the domain of both f and g.1. Sum of f and g: (f +g)(x) = f(x) + g(x)2. Difference of f and g: (f – g) (x) = f (x) - g (x)3. Product of f and g: (f • g) (x) = f(x) •g(x)

4. Quotient of f and g f(x)

(x) = , provided g(x) 0g(x)

f

g

Example 1

Let f(x) = x + 1 and g(x) = x2 -1 Find a rule for each of the following functions.

a. (f + g)(x) = f(x) + g(x) = (x + 1) +(x2 -1 ) = x2 + xb.

2

f(x) x + 1(x) = =

g(x) x 1

x + 1 =

(x +1)(x - 1)

1 = ,provided x 1

1

f

g

x

A fifth way of combining functions can be illustrated by the sport of cycling. When an 18 speed touring bicycle is in sixth gear. The gear ratio is 3:2 which means that the wheel of the bicycle revolve 3 times for every 2 revolutions of the pedal. this relationship can be expressed as

where w and p represent wheel and pedal revolutions, respectively. Since the wheels of a touring bicycle have a diameter of 27in. W revolutions of the wheels move the bicycle a distance d in inches, given by.

d = 27πw

3w = p

2

Notice that d = 27πw gives distance as a function of wheel revolutions and that w = 3 p 2Gives wheel revolutions as a functions of pedal revolutions. By substituting 3 p for w in d = 27πw 2We get d = 27π(3 p) = 40.5 πp 2Which gives distance as a function of wheel revolutions. This function of d(p) is said to be a composite of the functions d(w) and w(p).

The composite of a Function

The composite of f and g denoted f ⁰ g is denoted by two conditions:1. (f ⁰ g)(x) = f(g(x)) which is read “f circle g of x

equals f of g of x.2. X is the domain of g and g and g(x) is the

domain of f.

The domain of f ⁰ g is the set of x satisfying condition (2) above. The operation that combine f and g produce their composite is called the composite of functions.

Example 24 2

4 2

2

Let f(x) = x 3 d g(x) = x -2

( )( ) = f(g(x)) = f( x -2 )

= ( x -2 ) - 3( x -2 )

= x 7 10

x an

f g x

x

Example 31

f(x) = g(x) = x+1 find rules for x

( )( ) (g f)(x) and give the domain of

each composite function.

1( )( ) ( ( )) ( 1)

11 1

( )( ) ( ( )) 1

Let and

f g x and

f g x f g x f xx

g f x g f x gx x

HOMEWORK 1-6,17-22 PGS. 128-129

Class Example:1.) F(x) = 5 g(x) = 0 (f + g) (x) = 5 + 0 = 5b.) (f+g)(3) f(x) = 5 and g(x) = 3 (f+g) (x) = 5 + 3 = 8

Class Exercises`

2

2

3 2

2

2.) a.) 3 (b.) -3<x<3

5.)x 2 2 1

6.) 1

7.) x 2

8.) , for x -1

9.) a. 12

b. x 3 2

x

x

x x

x

x

Written Exercises3

3 3

5.) f(x) = x 1, ( ) 1

( )( ) x 1 1 x 2

g x x

f g x x x

3 36.) (f - g)(x) = x 1 1 =x - x x

37.) (f g)(x) = x 1( 1)x 4 3= x 1x x