Composite Function Worksheet - Fill Online, Printable, Fillable ... - Free Printable

Problem: Solve the composite function worksheet step by step.

#### Part 1: For \( f(x) = 2x - 1 \), \( g(x) = 3x \), and \( h(x) = x^2 + 1 \)

We need to compute the following:

1. \( f(g(-3)) \)
2. \( f(h(7)) \)
3. \( (g \circ h)(24) \)
4. \( f(g(h(2))) \)
5. \( h(f(f(5))) \)
6. \( g(f(h(-6))) \)
7. \( f(x + 1) \)
8. \( g(3a) \)
9. \( h(x - 2) \)

---

Solution for Part 1

#### 1. \( f(g(-3)) \)

- First, compute \( g(-3) \):
\[
g(x) = 3x \implies g(-3) = 3(-3) = -9
\]
- Next, compute \( f(g(-3)) = f(-9) \):
\[
f(x) = 2x - 1 \implies f(-9) = 2(-9) - 1 = -18 - 1 = -19
\]
- Answer: \( f(g(-3)) = -19 \)

#### 2. \( f(h(7)) \)

- First, compute \( h(7) \):
\[
h(x) = x^2 + 1 \implies h(7) = 7^2 + 1 = 49 + 1 = 50
\]
- Next, compute \( f(h(7)) = f(50) \):
\[
f(x) = 2x - 1 \implies f(50) = 2(50) - 1 = 100 - 1 = 99
\]
- Answer: \( f(h(7)) = 99 \)

#### 3. \( (g \circ h)(24) \)

- This is equivalent to \( g(h(24)) \).
- First, compute \( h(24) \):
\[
h(x) = x^2 + 1 \implies h(24) = 24^2 + 1 = 576 + 1 = 577
\]
- Next, compute \( g(h(24)) = g(577) \):
\[
g(x) = 3x \implies g(577) = 3(577) = 1731
\]
- Answer: \( (g \circ h)(24) = 1731 \)

#### 4. \( f(g(h(2))) \)

- First, compute \( h(2) \):
\[
h(x) = x^2 + 1 \implies h(2) = 2^2 + 1 = 4 + 1 = 5
\]
- Next, compute \( g(h(2)) = g(5) \):
\[
g(x) = 3x \implies g(5) = 3(5) = 15
\]
- Finally, compute \( f(g(h(2))) = f(15) \):
\[
f(x) = 2x - 1 \implies f(15) = 2(15) - 1 = 30 - 1 = 29
\]
- Answer: \( f(g(h(2))) = 29 \)

#### 5. \( h(f(f(5))) \)

- First, compute \( f(5) \):
\[
f(x) = 2x - 1 \implies f(5) = 2(5) - 1 = 10 - 1 = 9
\]
- Next, compute \( f(f(5)) = f(9) \):
\[
f(x) = 2x - 1 \implies f(9) = 2(9) - 1 = 18 - 1 = 17
\]
- Finally, compute \( h(f(f(5))) = h(17) \):
\[
h(x) = x^2 + 1 \implies h(17) = 17^2 + 1 = 289 + 1 = 290
\]
- Answer: \( h(f(f(5))) = 290 \)

#### 6. \( g(f(h(-6))) \)

- First, compute \( h(-6) \):
\[
h(x) = x^2 + 1 \implies h(-6) = (-6)^2 + 1 = 36 + 1 = 37
\]
- Next, compute \( f(h(-6)) = f(37) \):
\[
f(x) = 2x - 1 \implies f(37) = 2(37) - 1 = 74 - 1 = 73
\]
- Finally, compute \( g(f(h(-6))) = g(73) \):
\[
g(x) = 3x \implies g(73) = 3(73) = 219
\]
- Answer: \( g(f(h(-6))) = 219 \)

#### 7. \( f(x + 1) \)

- Substitute \( x + 1 \) into \( f(x) \):
\[
f(x) = 2x - 1 \implies f(x + 1) = 2(x + 1) - 1 = 2x + 2 - 1 = 2x + 1
\]
- Answer: \( f(x + 1) = 2x + 1 \)

#### 8. \( g(3a) \)

- Substitute \( 3a \) into \( g(x) \):
\[
g(x) = 3x \implies g(3a) = 3(3a) = 9a
\]
- Answer: \( g(3a) = 9a \)

#### 9. \( h(x - 2) \)

- Substitute \( x - 2 \) into \( h(x) \):
\[
h(x) = x^2 + 1 \implies h(x - 2) = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5
\]
- Answer: \( h(x - 2) = x^2 - 4x + 5 \)

---

Part 2: For \( f(x) = -3x + 7 \) and \( g(x) = 2x^2 - 8 \)

We need to compute the following:

10. \( f(g(x)) \)
11. \( (g \circ f)(x) \)

---

Solution for Part 2

#### 10. \( f(g(x)) \)

- Substitute \( g(x) \) into \( f(x) \):
\[
f(x) = -3x + 7 \quad \text{and} \quad g(x) = 2x^2 - 8
\]
\[
f(g(x)) = f(2x^2 - 8) = -3(2x^2 - 8) + 7 = -6x^2 + 24 + 7 = -6x^2 + 31
\]
- Answer: \( f(g(x)) = -6x^2 + 31 \)

#### 11. \( (g \circ f)(x) \)

- This is equivalent to \( g(f(x)) \).
- Substitute \( f(x) \) into \( g(x) \):
\[
f(x) = -3x + 7 \quad \text{and} \quad g(x) = 2x^2 - 8
\]
\[
g(f(x)) = g(-3x + 7) = 2(-3x + 7)^2 - 8
\]
- Compute \( (-3x + 7)^2 \):
\[
(-3x + 7)^2 = (-3x + 7)(-3x + 7) = 9x^2 - 42x + 49
\]
- Substitute back:
\[
g(f(x)) = 2(9x^2 - 42x + 49) - 8 = 18x^2 - 84x + 98 - 8 = 18x^2 - 84x + 90
\]
- Answer: \( (g \circ f)(x) = 18x^2 - 84x + 90 \)

---

Part 3: Additional Problems

#### 12. If \( f(x) = 3x - 5 \) and \( g(x) = x^2 \), find \( (f \circ g)(3) \)

- First, compute \( g(3) \):
\[
g(x) = x^2 \implies g(3) = 3^2 = 9
\]
- Next, compute \( f(g(3)) = f(9) \):
\[
f(x) = 3x - 5 \implies f(9) = 3(9) - 5 = 27 - 5 = 22
\]
- Answer: \( (f \circ g)(3) = 22 \)

#### 13. If \( f(x) = -9x - 9 \) and \( g(x) = \sqrt{x - 9} \), find \( (f \circ g)(10) \)

- First, compute \( g(10) \):
\[
g(x) = \sqrt{x - 9} \implies g(10) = \sqrt{10 - 9} = \sqrt{1} = 1
\]
- Next, compute \( f(g(10)) = f(1) \):
\[
f(x) = -9x - 9 \implies f(1) = -9(1) - 9 = -9 - 9 = -18
\]
- Answer: \( (f \circ g)(10) = -18 \)

---

Final Answers

\[
\boxed{
\begin{aligned}
1. & \ -19 \\
2. & \ 99 \\
3. & \ 1731 \\
4. & \ 29 \\
5. & \ 290 \\
6. & \ 219 \\
7. & \ 2x + 1 \\
8. & \ 9a \\
9. & \ x^2 - 4x + 5 \\
10. & \ -6x^2 + 31 \\
11. & \ 18x^2 - 84x + 90 \\
12. & \ 22 \\
13. & \ -18
\end{aligned}
}
\]