1. Operations with Functions
6-5
Operations with Functions
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt Algebra2

2. Simplify. Assume that all expressions are defined.
Warm Up
Simplify. Assume that all expressions are defined.
1. (2x + 5) – (x2 + 3x – 2)
–x2 – x + 7
2. (x – 3)(x + 1)2
x3 – x2 – 5x – 3 3.
x – 3
x – 2

3. Objectives Add, subtract, multiply, and divide functions.
Write and evaluate composite functions.

5. Example 1A: Adding and Subtracting Functions
Given f(x) = 4x2 + 3x – 1 and g(x) = 6x + 2, find each function.
(f + g)(x)
(f + g)(x) = f(x) + g(x)
= (4x2 + 3x – 1) + (6x + 2)
Substitute function rules.
= 4x2 + 9x + 1
Combine like terms.

6. Example 1B: Adding and Subtracting Functions
Given f(x) = 4x2 + 3x – 1 and g(x) = 6x + 2, find each function.
(f – g)(x)
(f – g)(x) = f(x) – g(x)
= (4x2 + 3x – 1) – (6x + 2)
Substitute function rules.
= 4x2 + 3x – 1 – 6x – 2
Distributive Property
= 4x2 – 3x – 3
Combine like terms.

7. Check It Out! Example 1a
Given f(x) = 5x – 6 and g(x) = x2 – 5x + 6, find each function.
(f + g)(x)
(f + g)(x) = f(x) + g(x)
= (5x – 6) + (x2 – 5x + 6)
Substitute function rules.
= x2
Combine like terms.

8. Check It Out! Example 1b
Given f(x) = 5x – 6 and g(x) = x2 – 5x + 6, find each function.
(f – g)(x)
(f – g)(x) = f(x) – g(x)
= (5x – 6) – (x2 – 5x + 6)
Substitute function rules.
= 5x – 6 – x2 + 5x – 6
Distributive Property
= –x2 + 10x – 12
Combine like terms.

9. When you divide functions, be sure to note any domain restrictions that may arise.

10. Example 2A: Multiplying and Dividing Functions
Given f(x) = 6x2 – x – 12 and g(x) = 2x – 3, find each function.
(fg)(x)
(fg)(x) = f(x) ● g(x)
Substitute function rules.
= (6x2 – x – 12) (2x – 3)
= 6x2 (2x – 3) – x(2x – 3) – 12(2x – 3)
Distributive Property
= 12x3 – 18x2 – 2x2 + 3x – 24x + 36
Multiply.
= 12x3 – 20x2 – 21x + 36
Combine like terms.

11. Example 2B: Multiplying and Dividing Functions
( )
(x) f g
( )
(x) f g
f(x)
g(x) =
6x2 – x –12
2x – 3 =
Set up the division as a rational expression. =
(2x – 3)(3x + 4)
2x – 3
Factor completely. Note that x ≠ 3 2 =
(2x – 3)(3x +4)
(2x – 3)
Divide out common factors.
= 3x + 4, where x ≠ 3 2
Simplify.

12. Another function operation uses the output from one function as the input for a second function. This operation is called the composition of functions.

13. The order of function operations is the same as the order of operations for numbers and expressions. To find f(g(3)), evaluate g(3) first and then substitute the result into f.

14. The composition (f o g)(x) or f(g(x)) is read “f of g of x.”
Reading Math

15. Be careful not to confuse the notation for multiplication of functions with composition fg(x) ≠ f(g(x))
Caution!

16. Example 3A: Evaluating Composite Functions
Given f(x) = 2x and g(x) = 7 – x, find each value.
f(g(4))
Step 1 Find g(4)
g(4) = 7 – 4
g(x) = 7 – x
= 3
Step 2 Find f(3)
f(3) = 23
f(x) = 2x
= 8
So f(g(4)) = 8.

17. Example 3B: Evaluating Composite Functions
Given f(x) = 2x and g(x) = 7 – x, find each value.
g(f(4))
Step 1 Find f(4)
f(4) = 24
f(x) = 2x
= 16
Step 2 Find g(16)
g(16) = 7 – 16
g(x) = 7 – x.
= –9
So g(f(4)) = –9.

18. You can use algebraic expressions as well as numbers as inputs into functions. To find a rule for f(g(x)), substitute the rule for g into f.

19. Example 4A: Writing Composite Functions
Given f(x) = x2 – 1 and g(x) = , write each composite function. State the domain of each. x
1 – x
f(g(x))
f(g(x)) = f( ) x
1 – x
Substitute the rule g into f.
= ( )2 – 1 x
1 – x
Use the rule for f. Note that x ≠ 1.
–1 + 2x
(1 – x)2 =
Simplify.
The domain of f(g(x)) is x ≠ 1 or {x|x ≠ 1} because g(1) is undefined.

20. Example 4B: Writing Composite Functions
Given f(x) = x2 – 1 and g(x) = , write each composite function. State the domain of each. x
1 – x
g(f(x))
g(f(x)) = g(x2 – 1)
Substitute the rule f into g.
(x2 – 1)
1 – (x2 – 1) =
Use the rule for g. =
x2 – 1
2 – x2
Simplify. Note that x ≠
The domain of g(f(x)) is x ≠ or {x|x ≠ } because f( ) = 1 and g(1) is undefined.

21. Composite functions can be used to simplify a series of functions.

22. Example 5: Business Application
Jake imports furniture from Mexico. The exchange rate is pesos per U.S. dollar. The cost of each piece of furniture is given in pesos. The total cost of each piece of furniture includes a 15% service charge.
A. Write a composite function to represent the total cost of a piece of furniture in dollars if the cost of the item is c pesos.

23. Step 1 Write a function for the total cost in U.S. dollars.
Example 5 Continued
Step 1 Write a function for the total cost in U.S. dollars.
P(c) = c c
= 1.15c
Step 2 Write a function for the cost in dollars based on the cost in pesos.
D(c) = c
11.30
Use the exchange rate.

24. Step 3 Find the composition D(P(c)).
Example 5 Continued
Step 3 Find the composition D(P(c)).
D(P(c)) = 1.15P(c)
Substitute P(c) for c.
= 1.15 ( ) c
11.30
Replace P(c) with its rule.
B. Find the total cost of a table in dollars if it costs 1800 pesos.
Evaluate the composite function for c = 1800.
D(P(c) ) = 1.15 ( )
1800
11.30 ≈
The table would cost $183.19, including all charges.

25. Lesson Quiz: Part I
Given f(x) = 4x2 – 1 and g(x) = 2x – 1, find each function or value.
1. (f + g)(x)
4x2 + 2x – 2
2. (fg)(x)
8x3 – 4x2 – 2x + 1
( )
(x) f g
2x + 1 3.
4. g(f(2)) 29

26. Lesson Quiz: Part II
Given f(x) = x2 and g(x) = , write each composite function. State the domain of each.
5. f(g(x))
f(g(x)) = x – 1;
{x|x ≥ 1}
6. g(f(x))
{x|x ≤ – 1 or x ≥ 1}