1. Digital Lesson
Algebra of Functions

2. Operations on Functions
If f and g are functions and x is an element of the domain of each function, then:
( f + g)(x) = f (x) + g(x)
( f – g)(x) = f (x) – g(x)
( f • g)(x) = f (x) • g(x)
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Operations on Functions

3. Example:(f+g)(x), (f-g)(x)
Example: Given f (x) = x2 and g(x) = 2, find ( f + g)(x) and ( f – g)(x).
( f + g)(x)
f(x) = x2
- 2 x y 2 4
( f + g)(x) = f(x) + g(x)
= x2 + 2
g(x) = 2
( f – g)(x) = f(x) – g(x)
= x2 – 2
( f – g)(x)
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Example:(f+g)(x), (f-g)(x)

4. Example: Given f (x) = 10 – 3x3 and g(x) = 2x2 + 1, find ( f + g)(2) and ( f – g)(–1) .
( f + g)(x) = (10 – 3x3) + (2x2 + 1)
= –3x3 + 2x2 + 11
( f + g)(2) = –3(2)3 + 2(2)2 + 11
= –5
(f – g)(x) = (10 – 3x3) – (2x2 + 1)
= –3x3 – 2x2 + 9
(f – g)(–1) = – 3(–1)3 – 2(–1)2 + 9
= 10
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Example: Sum of f and g

5. Example: Product of f and g
Example: Given f (x) = x – 2 and g(x) = x + 1, find ( f • g)(x) and ( f • g)(3) and ( f • g)(-1).
( f • g)(x) = (x – 2) • (x + 1)
- 2 x y 2 4
( f • g)(x)
= x2 – x – 2
g(x) = x + 1
( f • g)(3) = ((3) – 2) • ((3) + 1)
= (3)2 – (3) – 2
f(x) = x – 2
= 4
( f • g)(-1) = (-1)2 – (-1) – 2
= 0
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Example: Product of f and g

6. Example: Quotient of f and g
Example: Given f (x) = 2x2 – 3x + 1 and g(x) = x – 1, find (1) and (2).
f(x) = 2x2 – 3x + 1
(1) = 1 x y 2 -2 1
This is not defined, so (1) cannot be determined.
(2) 2
g(x) = x – 1
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Example: Quotient of f and g

7. A function can be evaluated at the value of another function.
Let f and g be two functions. Where g(x) is in the domain of f for all x in the domain of g. The composition of the two functions denoted as (f g), is the function whose value at x is given by (f g)(x) = f [g(x)].
Example: Given f (x) = 2x + 5 and g(x) = x2 + 1, find f [g(–2)].
g(–2) = (–2)2 + 1 = 5
f [g(–2)] = f (5) = 2(5) + 5 = 15
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Composition function

8. Example: Given f (x) = 3x2 + 2x – 48 and g(x) = 5x – 14, evaluate the following composite functions:
1. g[ f (4)]
f (4) = 3(4)2 + 2(4) – 48 = 8
g[ f (4)] = g (8) = 5(8) – 14 = 26
2. f [ g(3)]
g(3) = 5(3) – 14 = 1
f [ g (3)] = f (1) = 3(1)2 + 2(1) – 48 = – 43
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Example: f[g(-2)]

9. Example: Given f (x) = 3x – 2 and g(x) = x2 – 2x,
When evaluating compositions of functions, the order in which the functions are applied is important.
Example: Given f (x) = 3x – 2 and g(x) = x2 – 2x,
find ( f ° g)(x) and (g ° f )(x).
g(x) = x2 – 2x y
- 2 x -4
( f ° g)(x) = f [g(x)]
= 3(x2 – 2x) – 2
= 3x2 – 6x – 2
f (x) = 3x – 2
( f ° g)(x)
Example continued
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Example: f[g(x)]

10. Example Continued: g[f(x)]
g(x) = x2 – 2x y
- 2 x -4
( g ° f )(x) = g [ f (x)]
= (3x – 2)2 – 2(3x – 2)
= 9x2 – 18x + 8
f (x) = 3x – 2
( f ° g)(x) = 3x2 – 6x – 2
The graphs of (f ° g )(x) and (g ° f )(x) intersect at only two points. There are only two values for x for which (f ° g )(x) and (g ° f )(x) are equal. (f ° g )(x) and (g ° f )(x) are different functions.
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Example Continued: g[f(x)]