Composite Functions Worksheet with Answers | Exercises Algebra ... - Free Printable

Let’s solve each problem step by step. We’ll go one at a time, carefully checking our work.

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Problem 1:
Given:
f(x) = -3x + 2
g(x) = x² + 4x - 7

Find: f(-6) and g(0)

→ For f(-6):
Plug in x = -6 into f(x):
f(-6) = -3*(-6) + 2 = 18 + 2 = 20

→ For g(0):
Plug in x = 0 into g(x):
g(0) = (0)² + 4*(0) - 7 = 0 + 0 - 7 = -7

✔ So far: f(-6) = 20, g(0) = -7

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Problem 2:
Same functions:
f(x) = -3x + 2
g(x) = x² + 4x - 7

Find: f(5) and g(-2)

→ For f(5):
f(5) = -3*(5) + 2 = -15 + 2 = -13

→ For g(-2):
g(-2) = (-2)² + 4*(-2) - 7 = 4 - 8 - 7 = -4 - 7 = -11

✔ So: f(5) = -13, g(-2) = -11

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Problem 3:
Same functions again.

Find: f(a+1) and g(b-1)

→ For f(a+1):
Replace x with (a+1):
f(a+1) = -3*(a+1) + 2 = -3a - 3 + 2 = -3a -1

→ For g(b-1):
Replace x with (b-1):
g(b-1) = (b-1)² + 4*(b-1) - 7
First expand (b-1)² = b² - 2b + 1
Then 4*(b-1) = 4b - 4
So altogether:
= b² - 2b + 1 + 4b - 4 - 7
Combine like terms:
b² + (-2b + 4b) + (1 - 4 - 7) = b² + 2b - 10

✔ So: f(a+1) = -3a -1, g(b-1) = b² + 2b - 10

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Problem 4:
Given: h(x) = x³ - 2x + 5
k(x) = √(x + 4)

Find: h(2) and k(5)

→ For h(2):
h(2) = (2)³ - 2*(2) + 5 = 8 - 4 + 5 = 4 + 5 = 9

→ For k(5):
k(5) = √(5 + 4) = √9 = 3

✔ So: h(2) = 9, k(5) = 3

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Problem 5:
Same functions: h(x) = x³ - 2x + 5, k(x) = √(x + 4)

Find: h(-1) and k(0)

→ For h(-1):
h(-1) = (-1)³ - 2*(-1) + 5 = -1 + 2 + 5 = 1 + 5 = 6

→ For k(0):
k(0) = √(0 + 4) = √4 = 2

✔ So: h(-1) = 6, k(0) = 2

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Problem 6:
Same functions.

Find: h(c+2) and k(d-3)

→ For h(c+2):
Replace x with (c+2):
h(c+2) = (c+2)³ - 2*(c+2) + 5

First, expand (c+2)³:
= c³ + 3*c²*2 + 3*c*4 + 8 = c³ + 6c² + 12c + 8
Wait — let me do it properly:

Actually, (c+2)³ = c³ + 3c²(2) + 3c(2²) + 2³ = c³ + 6c² + 12c + 8

Now subtract 2*(c+2) = 2c + 4 → so minus that is -2c -4

Then add 5.

So total:
c³ + 6c² + 12c + 8 - 2c - 4 + 5
Combine like terms:
c³ + 6c² + (12c - 2c) + (8 - 4 + 5) = c³ + 6c² + 10c + 9

→ For k(d-3):
k(d-3) = √((d-3) + 4) = √(d + 1)

✔ So: h(c+2) = c³ + 6c² + 10c + 9, k(d-3) = √(d + 1)

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Problem 7:
Given: p(x) = 2x² - 3x + 1
q(x) = |x - 5|

Find: p(3) and q(8)

→ For p(3):
p(3) = 2*(3)² - 3*(3) + 1 = 2*9 - 9 + 1 = 18 - 9 + 1 = 9 + 1 = 10

→ For q(8):
q(8) = |8 - 5| = |3| = 3

✔ So: p(3) = 10, q(8) = 3

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Problem 8:
Same functions.

Find: p(-2) and q(2)

→ For p(-2):
p(-2) = 2*(-2)² - 3*(-2) + 1 = 2*4 + 6 + 1 = 8 + 6 + 1 = 15

→ For q(2):
q(2) = |2 - 5| = |-3| = 3

✔ So: p(-2) = 15, q(2) = 3

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Problem 9:
Same functions.

Find: p(m+1) and q(n-4)

→ For p(m+1):
p(m+1) = 2*(m+1)² - 3*(m+1) + 1
First, (m+1)² = m² + 2m + 1
So 2*(that) = 2m² + 4m + 2
Then -3*(m+1) = -3m -3
Add 1.

Total:
2m² + 4m + 2 - 3m - 3 + 1
Combine:
2m² + (4m - 3m) + (2 - 3 + 1) = 2m² + m + 0 = 2m² + m

→ For q(n-4):
q(n-4) = |(n-4) - 5| = |n - 9|

✔ So: p(m+1) = 2m² + m, q(n-4) = |n - 9|

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Problem 10:
Given: r(x) = 1/(x+3), s(x) = √(x² - 4)

Find: r(2) and s(3)

→ For r(2):
r(2) = 1/(2+3) = 1/5 = 0.2 or leave as fraction: 1/5

→ For s(3):
s(3) = √(3² - 4) = √(9 - 4) = √5 ≈ but we can leave exact: √5

✔ So: r(2) = 1/5, s(3) = √5

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Problem 11:
Same functions.

Find: r(-1) and s(2)

→ For r(-1):
r(-1) = 1/(-1 + 3) = 1/2 = 0.5 or 1/2

→ For s(2):
s(2) = √(2² - 4) = √(4 - 4) = √0 = 0

✔ So: r(-1) = 1/2, s(2) = 0

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Problem 12:
Same functions.

Find: r(t+1) and s(u+2)

→ For r(t+1):
r(t+1) = 1 / ((t+1) + 3) = 1 / (t + 4)

→ For s(u+2):
s(u+2) = √[(u+2)² - 4]
Expand (u+2)² = u² + 4u + 4
Subtract 4: u² + 4u + 4 - 4 = u² + 4u
So s(u+2) = √(u² + 4u)

We can factor inside: √[u(u + 4)] — but unless asked to simplify further, this is fine.

✔ So: r(t+1) = 1/(t+4), s(u+2) = √(u² + 4u)

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Now, let’s compile all answers clearly.

Final Answer:
1. f(-6) = 20, g(0) = -7
2. f(5) = -13, g(-2) = -11
3. f(a+1) = -3a - 1, g(b-1) = b² + 2b - 10
4. h(2) = 9, k(5) = 3
5. h(-1) = 6, k(0) = 2
6. h(c+2) = c³ + 6c² + 10c + 9, k(d-3) = √(d + 1)
7. p(3) = 10, q(8) = 3
8. p(-2) = 15, q(2) = 3
9. p(m+1) = 2m² + m, q(n-4) = |n - 9|
10. r(2) = 1/5, s(3) = √5
11. r(-1) = 1/2, s(2) = 0
12. r(t+1) = 1/(t+4), s(u+2) = √(u² + 4u)