7.4 Composition of Functions

7.4 Composition of Functions

2/26/2014

Function:

A function is a special relationship between values: Each of its input values gives back exactly one output value.

It is often written as "f(x)" where x is the input value.

Ex 1: Gfind: a.) Solution: Substitute 0 for x in 3x+ 1

3(0) + 1 = 1 = 1

b.) 3(½) + 1 = 1 + 1 = 2

c.) 3(a) + 1 = 3a + 1

d.) 3(a+3) + 1 = 3a + 9 +1 = 3a+10

e.) = 3a + 4

Ex 2: Gfind: a.)

= -5

b.)

= -c.)

=

d.)

= e.)

3 =

f

g

y

x

2

4

6

8

2 4 6 8–2–4–6–8

–2

–4Use the graph to find:a. f(0) Solution: at x = 0, the graph of the f (red) function is at 2.f(0) = 2b. g(0)Solution: at x = 0, the graph of the g (black) function is at 3g(0) = 3

Composition of functionsoccurs when you insert one function into another. In effect, the range (output) of the inside function becomes the domain (input) of the outside function. The notation for composition of functions is either

Or

Note: which means order matters!

f

g

y

x

2

4

6

8

2 4 6 8–2–4–6–8

–2

–4Use the graph to find:c.) f(g(0))Solution: Evaluate g(0) first, and from the previous problem (b), g(0) = 3. Then look at the graph of the f function and see what the y component when x = 3.At x = 3 the red graph is at -1f(g(0)) = -1

f

g

y

x

2

4

6

8

2 4 6 8–2–4–6–8

–2

–4Use the graph to find:d.) g(f(0))Solution: Evaluate f(0) which in problem (a) is 2. Then look at the graph of the g function and see what the y component when x = 2.At x = 2 the black graph is at 5.g(f(0)) = 5

f

g

y

x

2

4

6

8

2 4 6 8–2–4–6–8

–2

–4

Use the graph to find:a.) f(g(-1)) and b.) g (f(-1)) a.) Solution: Evaluate g(-1)g(-1) = 2 then evaluate f(2)f(2) = 0b.) Solution: Evaluate f(-1)f(-1) = 3 then evaluate g(3)g(3) = 6

Given:

Evaluate the following expressions:a. f(-1)b. g(-1)c. f(g(-1))d. g(f(-1))

a.)

b.)

c.)

d.)

Homework:WS 7.4 #1-9

Example 1 Add and Subtract Functions

Let and . Find: =f ( )x 4x 2 =g( )x x 1+

f ( )x g( )x+a. b. f ( )x g( )x–

SOLUTION

=h( )x f ( )x g( )x+a. f ( )x g( )xb. –

4x 2 ( )1x ++= 4x 2 ( )1x +–

In both parts and , the domains of f and g are all real numbers. So, the domain of h is all real numbers.

( )a ( )b

4x 2 1x ++= 4x 2 1x– –

Let and . Find:

Example 2 Multiply and Divide Functions

=f ( )x x 3 =g( )x 2x

f ( )x g( )xa. • b.f ( )x

g( )x

=2x 4

=21

x 2

𝑓 (𝑥 ) ∙𝑔 (𝑥 )=𝑥3 ∙2𝑥 𝑓 (𝑥 )𝑔 (𝑥 )

= 𝑥3

2𝑥

Checkpoint Perform Function Operations

1.

Let and . Find=f ( )x 3x =g( )x x 1–

f ( )x g( )x+ANSWER

4x 1–

3x 2 3x,–2. f ( )x g( )x•

3. f ( )x

g( )x

3x

x 1–, x 1=