more language of functions x

1. Notation and Algebra of Functions

2. Functions are procedures that assign a unique output to each
(valid) input.
Notation and Algebra of Functions

3. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions

4. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
name of
the function
the input box the defining
formula

5. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input box “( )” holds the input to be evaluated by the
formula.

6. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.

7. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.

8. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.

9. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
(The Square–Formula) (a ± b)2 = a2 ± 2ab + b2
We write down the Square–Formula as a reminder below.

10. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
(The Square–Formula) (a ± b)2 = a2 ± 2ab + b2
We write down the Square–Formula as a reminder below.
Example A.
Given f(x) = x2 – 2x + 3, simplify the following.
a. f (2a) =
b. f (a + b) =

11. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
(The Square–Formula) (a ± b)2 = a2 ± 2ab + b2
We write down the Square–Formula as a reminder below.
Example A.
Given f(x) = x2 – 2x + 3, simplify the following.
a. f (2a) = (2a)2 – 2(2a) + 3
b. f (a + b) =

12. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
(The Square–Formula) (a ± b)2 = a2 ± 2ab + b2
We write down the Square–Formula as a reminder below.
Example A.
Given f(x) = x2 – 2x + 3, simplify the following.
a. f (2a) = (2a)2 – 2(2a) + 3
= 4a2 – 4a + 3
b. f (a + b) =

13. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
(The Square–Formula) (a ± b)2 = a2 ± 2ab + b2
We write down the Square–Formula as a reminder below.
Example A.
Given f(x) = x2 – 2x + 3, simplify the following.
a. f (2a) = (2a)2 – 2(2a) + 3
= 4a2 – 4a + 3
b. f (a + b) = (a + b)2 – 2(a + b) + 3

14. Functions are procedures that assign a unique output to each
(valid) input. Most mathematics functions are given in
mathematics formulas such as f (x) = x2 – 2x + 3 = y.
Notation and Algebra of Functions
The input may be other mathematics expressions. Often in
such problems we are to simplify the outputs algebraically.
The input box “( )” holds the input to be evaluated by the
formula. Hence f(2) means to replace x by the input (2) in the
formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
(The Square–Formula) (a ± b)2 = a2 ± 2ab + b2
We write down the Square–Formula as a reminder below.
Example A.
Given f(x) = x2 – 2x + 3, simplify the following.
a. f (2a) = (2a)2 – 2(2a) + 3
= 4a2 – 4a + 3
b. f (a + b) = (a + b)2 – 2(a + b) + 3
= a2 + 2ab + b2 – 2a – 2b + 3

15. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions

16. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.

17. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3

18. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7

19. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)

20. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1

21. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3)

22. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12

23. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)

24. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.

25. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)

26. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3

27. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3
= 4 – 4 – 3

28. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3
= 4 – 4 – 3
= –3

29. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3
= 4 – 4 – 3
= –3
g(–3) = 2(–3) +1

30. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3
= 4 – 4 – 3
= –3
g(–3) = 2(–3) +1
= –6 + 1
= –5

31. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3
= 4 – 4 – 3
= –3
Hence f(2) – g(-3)
= –3 – (–5)
g(–3) = 2(–3) +1
= –6 + 1
= –5

32. We may form new functions by +, – , * , and / functions.
Notation and Algebra of Functions
Example B. Let f(x) = 3x – 4, g(x) = x2 – 2x – 3, find f + g, f – g,
f*g, and f/g.
f + g = 3x – 4 + x2 – 2x – 3 = x2 + x – 7
f – g = 3x – 4 – (x2 – 2x – 3)
= 3x – 4 – x2 + 2x + 3 = –x2 + 5x – 1
f*g = (3x – 4)*(x2 – 2x – 3) = 3x3 – 10x2 – x + 12
f/g = (3x – 4)/(x2 – 2x – 3)
Algebraic expressions may be formed with functions notations.
Example C. Let f(x) = x2 – 2x – 3, g(x) = 2x + 1.
a. Simplify f(2) – g(–3)
We find f(2) = (2)2 – 2*(2) – 3
= 4 – 4 – 3
= –3
Hence f(2) – g(-3)
= –3 – (–5) = –3 + 5 = 2
g(–3) = 2(–3) +1
= –6 + 1
= –5

33. b. Simplify f(x + h ) – f(x)
Notation and Algebra of Functions

34. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h )
Notation and Algebra of Functions

35. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
Notation and Algebra of Functions

36. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Notation and Algebra of Functions

37. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
Notation and Algebra of Functions

38. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
Notation and Algebra of Functions

39. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions

40. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions
Composition of Functions

41. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions
Composition of Functions
When one function is used as the input of another function,
this operation is called composition.

42. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions
Composition of Functions
When one function is used as the input of another function,
this operation is called composition.
Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).

43. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions
Composition of Functions
When one function is used as the input of another function,
this operation is called composition.
Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).
(It's read as f "circled" g).

44. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions
Composition of Functions
When one function is used as the input of another function,
this operation is called composition.
Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).
(It's read as f "circled" g).
In other words, the input for f is g(x).

45. b. Simplify f(x + h ) – f(x)
First, we calculate
f(x + h ) = (x+h)2 – 2(x+h) – 3
= x2 + 2xh + h2 – 2x – 2h – 3
Hence f(x+h) – f(x)
= x2 + 2xh + h2 – 2x – 2h – 3 – (x2 – 2x – 3)
= x2 + 2xh + h2 – 2x – 2h – 3 – x2 + 2x + 3
= 2xh + h2 – 2h
Notation and Algebra of Functions
Composition of Functions
When one function is used as the input of another function,
this operation is called composition.
Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).
(It's read as f "circled" g).
In other words, the input for f is g(x).
Similarly, we define (g ○ f)(x) ≡ g (f(x)).

46. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
Notation and Algebra of Functions

47. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Notation and Algebra of Functions

48. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3
Notation and Algebra of Functions

49. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Notation and Algebra of Functions

50. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9)
Notation and Algebra of Functions

51. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
Notation and Algebra of Functions

52. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Notation and Algebra of Functions

53. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Since f(3) = 3(3) – 5
Notation and Algebra of Functions

54. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Since f(3) = 3(3) – 5 = 4
Notation and Algebra of Functions

55. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Since f(3) = 3(3) – 5 = 4
Hence g(f(3)) =g(4)
Notation and Algebra of Functions

56. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Since f(3) = 3(3) – 5 = 4
Hence g(f(3)) =g(4) = –4(4) + 3
Notation and Algebra of Functions

57. Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Since f(3) = 3(3) – 5 = 4
Hence g(f(3)) =g(4) = –4(4) + 3 = –13
Notation and Algebra of Functions

58. Note: They are different!
Example C.
Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3).
(f ○ g)(3) = f(g(3))
Since g(3) = –4(3) + 3 = –9
Hence f(g(3)) = f(–9) = 3(–9) – 5 = –32
(g ○ f)(3) = g(f(3))
Since f(3) = 3(3) – 5 = 4
Hence g(f(3)) =g(4) = –4(4) + 3 = –13
Notation and Algebra of Functions

59. f(x) = 2 – 3x g(x) = –2x2 + 3x – 1 h(x) =
Exercise. A. Simplify the following expressions
with the given functions.
Notation and Algebra of Functions
2x – 1
x – 2
8. f(2a)
12. 2h(a)
9. g(2a) 11. h(2a)
10. 2g(a)
13. f(3 + b) 14. g(3 + b) 15. h(3 + b)
1. f(2) + f(3)
5. f(0) + g(0) + h(0)
2. 2f(3) 4. [h(2)]2
3. 2g(0) + g(1)
6. [f(3)]2 – [g(3)]2 7. h(1) / h(–1 )
16. f(3 + b) – f(b) 17. g(3 + b) – g(b) 18. h(3 + b) – h(b)
19. f(3 + b) – f(3 – b) 20. g(3 + b) – g(3 – b)
24. (f ○ g)(2) 25. (g ○ f)(–2)
Exercise. B. Simplify the following compositions.
26. (g ○ g)(–1) 27. (h ○ g)(3)
28. (f ○ h)(1) 29. (f ○ f)(1) 30. (f ○ g)(x) 31. (g ○ f)(x)
32. (f ○ f)(x) 33. (h ○ f)(x) 34. (f ○ h)(x)
21. g(x) + 3f(x) 22. 2g(x) + [f(x)]2 23. g(x) / h(x)