π What is a Composite Function?
A composite function is formed when the output of one function is used as the input of another. We evaluate the inner function first, then plug that result into the outer function.
(f β g)(x) = f(g(x))
Both notations mean the same thing: evaluate g first, then plug into f.
Read aloud: βf of g of xβ
Key Rule: When evaluating f(g(x)), always work inside-out (right to left). Evaluate g(x) first, then use that answer as the input for f.
βοΈ Warm-Up β Substituting Expressions
f(x) = 2x β 3
f(5)
= 2(5) β 3 = 7
f(a)
= 2a β 3
f(4t)
= 2(4t) β 3 = 8t β 3
f(5x β 1)
= 2(5x β 1) β 3
= 10x β 2 β 3
= 10x β 5
βοΈ Example 1 β Numeric Composition
f(x) = 3x + 1. Let f(2) = b. Find f(b).
1
Find b: f(2) = 3(2) + 1 = 7 β b = 7
2
Find f(b): f(7) = 3(7) + 1 = 22
βοΈ Example 2 β Evaluating Composites at a Number
f(x) = 3x β 5 Β· g(x) = 2x + 1
f(g(3))
1
g(3) = 2(3)+1 = 7
2
f(7) = 3(7)β5 = 16
g(f(3))
1
f(3) = 3(3)β5 = 4
2
g(4) = 2(4)+1 = 9
f(f(4))
1
f(4) = 3(4)β5 = 7
2
f(7) = 3(7)β5 = 16
Notice: f(g(3)) β g(f(3)). Composition is generally NOT commutative β order matters!
βοΈ Example 3 β Composition as an Expression
f(x) = 3x β 5 Β· g(x) = 2x + 1
f(g(x))
f(g(x)) = f(2x+1)
= 3(2x+1) β 5
= 6x+3β5
= 6x β 2
= 3(2x+1) β 5
= 6x+3β5
= 6x β 2
g(f(x))
g(f(x)) = g(3xβ5)
= 2(3xβ5) + 1
= 6xβ10+1
= 6x β 9
= 2(3xβ5) + 1
= 6xβ10+1
= 6x β 9
βοΈ Example 4 β Mixed Function Types
h(x) = 2xβ3 Β· k(x) = xΒ²+4x+5 Β· p(x) = β(3x+1) Β· m(x) = 3x+2
k(h(2))
1
h(2) = 2(2)β3 = 1
2
k(1) = 1+4+5 = 10
(h β m)(β1)
1
m(β1) = 3(β1)+2 = β1
2
h(β1) = 2(β1)β3 = β5
(k β p)(x)
k(p(x)) = k(β(3x+1))
= (β(3x+1))Β² + 4β(3x+1) + 5
= 3x+1 + 4β(3x+1) + 5
= 3x+6+4β(3x+1)
= (β(3x+1))Β² + 4β(3x+1) + 5
= 3x+1 + 4β(3x+1) + 5
= 3x+6+4β(3x+1)
βοΈ Example 5 β Tables, Graphs, Piecewise & Verbal
Multiple Representations
f(x) from graph Β· g(x) from table Β· h(x) piecewise Β· k(x) = (xβ2)Β²β3 (verbal: quadratic, shift right 2, down 3)
| x | β4 | β1 | 2 | 5 | 7 |
|---|---|---|---|---|---|
| g(x) | Ο | 3 | 3 | β2 | 4 |
h(x) = |β7+3x|, x<1
β4, x=1
βx, x>1
β4, x=1
βx, x>1
f(g(3))
g(3) = not defined (3 not in table)
β not defined
(hβf)(8)
1
f(8) = β1 (from graph)
2
h(β1) = |β7+3(β1)| = |β10| = 10
(gβk)(0)
1
k(0) = (0β2)Β²β3 = 4β3 = 1
2
g(1) = not defined β not defined
f(f(2))
1
f(2) = 1 (from graph)
2
f(1) = 4 (from graph)
(gβh)(1)
1
h(1) = β4 (x=1 case)
2
g(β4) = Ο (from table)
(gβfβhβk)(3)
1
k(3) = (3β2)Β²β3 = β2
2
h(β2) = |β7+3(β2)| = 13
3
f(13) = not defined
Key takeaway: A composite function is not defined whenever any intermediate step produces a value outside the domain of the next function.
Card 1 of 6
Concept
What does f(g(x)) mean? How do you evaluate it?
Tap to reveal β¨
Answer
Evaluate g(x) FIRST, then plug that result into f
Work inside-out (right to left). f(g(x)) = f applied to the output of g. Also written (f β g)(x).
Topic 2.7
Composition of Functions
0 / 5
Question 1 of 5
f(x) = 3x β 5 and g(x) = 2x + 1. Evaluate f(g(3)).
Start with the inner function: g(3) = ?
Question 2 of 5
f(x) = 3x β 5 and g(x) = 2x + 1. Find an expression for f(g(x)).
Substitute g(x) = 2x + 1 into f
Question 3 of 5
h(x) = 2x β 3 and k(x) = xΒ² + 4x + 5. Evaluate k(h(2)).
h(2) = 2(2) β 3 = 1, then k(1) = ?
Question 4 of 5
Using the table: g(β4)=Ο, g(β1)=3, g(2)=3, g(5)=β2, g(7)=4. And h(x): h(1)=β4, h(x)=βx for x>1. Evaluate (g β h)(1).
Start with h(1), then plug into g
Question 5 of 5
f(x) = 3x β 5, g(x) = 2x + 1. Is f(g(x)) = g(f(x))? Justify.
f(g(x)) = 6x β 2 Β· g(f(x)) = 6x β 9
0/5
Keep going!