Composition of Functions worksheet with ten problems requiring students to find values of composite functions.

Problem: Composition of Functions

The task involves performing the indicated operations for each pair of functions. Specifically, we need to evaluate compositions of functions such as \( f(g(x)) \), \( h(f(x)) \), etc.

#### Step-by-Step Solutions:

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1. \( f(x) = 4x - 7 \) and \( y(x) = 3x^2 + 1 \)

Find: \( f(y(-1)) \)

#### Step 1: Evaluate \( y(-1) \)
\[
y(x) = 3x^2 + 1
\]
\[
y(-1) = 3(-1)^2 + 1 = 3(1) + 1 = 3 + 1 = 4
\]

#### Step 2: Substitute \( y(-1) = 4 \) into \( f(x) \)
\[
f(x) = 4x - 7
\]
\[
f(y(-1)) = f(4) = 4(4) - 7 = 16 - 7 = 9
\]

Answer:
\[
\boxed{9}
\]

---

2. \( f(x) = x^2 + 3x \) and \( h(x) = 4x \)

Find: \( f(h(-2)) \)

#### Step 1: Evaluate \( h(-2) \)
\[
h(x) = 4x
\]
\[
h(-2) = 4(-2) = -8
\]

#### Step 2: Substitute \( h(-2) = -8 \) into \( f(x) \)
\[
f(x) = x^2 + 3x
\]
\[
f(h(-2)) = f(-8) = (-8)^2 + 3(-8) = 64 - 24 = 40
\]

Answer:
\[
\boxed{40}
\]

---

3. \( g(t) = -2t + 4 \) and \( f(t) = t + 5 \)

Find: \( g(f(3)) \)

#### Step 1: Evaluate \( f(3) \)
\[
f(t) = t + 5
\]
\[
f(3) = 3 + 5 = 8
\]

#### Step 2: Substitute \( f(3) = 8 \) into \( g(t) \)
\[
g(t) = -2t + 4
\]
\[
g(f(3)) = g(8) = -2(8) + 4 = -16 + 4 = -12
\]

Answer:
\[
\boxed{-12}
\]

---

4. \( h(x) = x^2 + 2x \) and \( f(x) = x - 7 \)

Find: \( h(f(1)) \)

#### Step 1: Evaluate \( f(1) \)
\[
f(x) = x - 7
\]
\[
f(1) = 1 - 7 = -6
\]

#### Step 2: Substitute \( f(1) = -6 \) into \( h(x) \)
\[
h(x) = x^2 + 2x
\]
\[
h(f(1)) = h(-6) = (-6)^2 + 2(-6) = 36 - 12 = 24
\]

Answer:
\[
\boxed{24}
\]

---

5. \( f(x) = 2x^2 + x \) and \( y(x) = x + 2 \)

Find: \( f(y(-5)) \)

#### Step 1: Evaluate \( y(-5) \)
\[
y(x) = x + 2
\]
\[
y(-5) = -5 + 2 = -3
\]

#### Step 2: Substitute \( y(-5) = -3 \) into \( f(x) \)
\[
f(x) = 2x^2 + x
\]
\[
f(y(-5)) = f(-3) = 2(-3)^2 + (-3) = 2(9) - 3 = 18 - 3 = 15
\]

Answer:
\[
\boxed{15}
\]

---

6. \( y(x) = x^3 - 5x \)

Find: \( y(y(-1)) \)

#### Step 1: Evaluate \( y(-1) \)
\[
y(x) = x^3 - 5x
\]
\[
y(-1) = (-1)^3 - 5(-1) = -1 + 5 = 4
\]

#### Step 2: Substitute \( y(-1) = 4 \) into \( y(x) \)
\[
y(x) = x^3 - 5x
\]
\[
y(y(-1)) = y(4) = (4)^3 - 5(4) = 64 - 20 = 44
\]

Answer:
\[
\boxed{44}
\]

---

7. \( g(x) = 4x^2 - 2 \) and \( y(x) = 2x \)

Find: \( g(y(-3)) \)

#### Step 1: Evaluate \( y(-3) \)
\[
y(x) = 2x
\]
\[
y(-3) = 2(-3) = -6
\]

#### Step 2: Substitute \( y(-3) = -6 \) into \( g(x) \)
\[
g(x) = 4x^2 - 2
\]
\[
g(y(-3)) = g(-6) = 4(-6)^2 - 2 = 4(36) - 2 = 144 - 2 = 142
\]

Answer:
\[
\boxed{142}
\]

---

8. \( f(t) = t^2 - 5 \) and \( y(t) = t + 9 \)

Find: \( f(y(-4)) \)

#### Step 1: Evaluate \( y(-4) \)
\[
y(t) = t + 9
\]
\[
y(-4) = -4 + 9 = 5
\]

#### Step 2: Substitute \( y(-4) = 5 \) into \( f(t) \)
\[
f(t) = t^2 - 5
\]
\[
f(y(-4)) = f(5) = (5)^2 - 5 = 25 - 5 = 20
\]

Answer:
\[
\boxed{20}
\]

---

9. \( f(x) = 5x^3 \) and \( y(x) = -x + 2 \)

Find: \( f(y(2)) \)

#### Step 1: Evaluate \( y(2) \)
\[
y(x) = -x + 2
\]
\[
y(2) = -(2) + 2 = -2 + 2 = 0
\]

#### Step 2: Substitute \( y(2) = 0 \) into \( f(x) \)
\[
f(x) = 5x^3
\]
\[
f(y(2)) = f(0) = 5(0)^3 = 5(0) = 0
\]

Answer:
\[
\boxed{0}
\]

---

10. \( y(x) = x^3 - 10 \) and \( h(x) = x - 2 \)

Find: \( y(h(2)) \)

#### Step 1: Evaluate \( h(2) \)
\[
h(x) = x - 2
\]
\[
h(2) = 2 - 2 = 0
\]

#### Step 2: Substitute \( h(2) = 0 \) into \( y(x) \)
\[
y(x) = x^3 - 10
\]
\[
y(h(2)) = y(0) = (0)^3 - 10 = 0 - 10 = -10
\]

Answer:
\[
\boxed{-10}
\]

---

Final Answers:

1. \( \boxed{9} \)
2. \( \boxed{40} \)
3. \( \boxed{-12} \)
4. \( \boxed{24} \)
5. \( \boxed{15} \)
6. \( \boxed{44} \)
7. \( \boxed{142} \)
8. \( \boxed{20} \)
9. \( \boxed{0} \)
10. \( \boxed{-10} \)