Operations on Functions

Section 1-8

Objectives

I can add, subtract, multiply, and divide functions

I can find the domains of newly formed functions

Chapter 1 Section 8

Arithmetic of Functions

NOTATION

f(x) + g(x) means

Add the f(x) function to the g(x) function.

Chapter 1 Section 8

Arithmetic of Functions

NOTATION

f(x) - g(x) means

Subtract the g(x) function from the f(x) function.

Chapter 1 Section 8

Arithmetic of Functions

NOTATION

(fg)(x) means

f(x) g(x)

Multiply f(x) times g(x)

Chapter 1 Section 8

Arithmetic of Functions

NOTATION

(f/g)(x) means

f(x) g(x)

Divide f(x) by g(x)

Let f(x) = 3x + 4 & g(x) = 2x – 3find:a) f(x) + g(x)

b) f(x) – g(x) c) f(x)g(x)

d) f(x)/g(x)

5x + 1

x + 7

6x2 – x - 12

3 4 3;

2 3 2

xx

x

Operations on Functions:

Addition: (f + g)(x) = f(x) + g(x)

Example: If f(x) = 3x – 6 and g(x) = -3x2 + 3x + 4 then

(f + g)(x) = 3x –6 + -3x2 + 3x + 4 = -3x2 + 6x –2

Subtraction: (f – g)(x) = f(x) – g(x)

Example: Use the same functions as above. Then

(f - g)(x) = 3x – 6 – (-3x2 + 3x + 4 )

(f – g)(x) = 3x2 - 10

Operations on Functions:

Multiplication: (fg)(x) = f(x)g(x)

Example: If f(x) = 3x – 6 and g(x) = -3x2 + 3x + 4 then

(fg)(x) = (3x –6)(-3x2 + 3x + 4) = -9x3 + 9x2 + 12x + 18x2 – 18x – 24 = -9x3 + 27x2 – 6x - 24

Division: (f/g)(x) = f(x)/g(x)

Example: Use the same functions as above. Then

(f/g)(x) = (3x – 6)/(-3x2 + 3x + 4) except for x – values that make denominator 0

Finding the domains of our new functions

Finding the domains of these new functions is a little complicated.

The first step is to find the domain of each function.

The second step is to find the intersection of the domains. This intersection is the domain of the new function*.

* Unless the operation is division in which case you must also exclude x-values that make the denominator zero

If f and g are functions with domainsA and B:

Their sum f + g is the function givenby

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

Domain of f + g

BA

Lets do an example; Let

a) Find (f + g)(x) and b) it’s domain.

(f + g)(x) =

xxgxxf )( and 2)(

xx 2

Step 1: The domain of f(x) is [-2,). The domain of g(x) is [0,).

Step 2: [-2,) [0,) =

[0,).

Since this is not a quotient I don’t have to worry about division by zero.

Their difference f - g is thefunction given by

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

Domain of f - g

BA

Lets do an example; Let

a) Find (f - g)(x) and b) it’s domain.

(f - g)(x) =

4)( and 23)( 2 xxgxxf

232 xx

Step 1: The domain of f(x) is (- ,). The domain of g(x) is (- ,).

Step 2: (- ,) (- ,) =

(- ,).

Since this is not a quotient I don’t have to worry about division by zero.

Their product f g is thefunction given by

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

Domain of f g

BA

Example; Let

a) Find (f · g)(x) and b) it’s domain.

(f · g)(x) =

4)( and 23)( xxgxxf

8103 2 xx

Step 1: The domain of f(x) is (- ,). The domain of g(x) is (- ,).

Step 2: (- ,) (- ,) =

(- ,).

Since this is not a quotient I don’t have to worry about division by zero.

Their quotient f / g is thefunction given by

( f / g )(x ) = f(x ) / g (x ) , g (x ) 0

Domain of f / g

0)( xgBA

Another example:

10

)(

x

xx

g

f

a) Find (f /g)(x) and b) it’s domain.

Step 1: Domain of f(x) is [0,). Domain of g(x) is [10, )

Step 2: [0,) [10, ) = [10, )*

* Division so we can’t plug in 10 so Domain is (10, ).

10)( and )( xxgxxf

Example

Given f(x) = x – 5 and g(x) = x2 -1, find (f+g)(x), (f - g)(x), (fg)(x) and (f/g)(x)

(f+g)(x) = f(x) + g(x) = (x – 5)+(x2 –1) =x2 + x - 6 Domain (f+g)(x): (- , )

(f – g)(x) = f(x) – g(x) = (x – 5) – (x2 –1 ) = x – 5 – x2 + 1 = -x2 + x – 4 Domain (f – g)(x) = (- , )

Example

Given f(x) = x – 5 and g(x) = x2 -1, find (f - g)(x), (fg)(x) and (f/g)(x)

(fg)(x) = f(x)g(x) = (x – 5)(x2 –1) = x3 – x – 5x2 + 5 = x3 – 5x2 – x + 5 Domain (fg)(x) : (- , )

(f/g)(x)=f(x)/g(x) = (x – 5)/(x2 – 1) Domain (f/g)(x): {All real except x 1 or –1} (- , -1) U (-1,1) U (1, )

Homework

WS 2-1