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Problem: Composition of Functions - Worksheet 1

We are tasked with solving several problems involving function composition. Let's break down each part step by step.

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#### Part 1: Using \( f(x) = 4x + 3 \) and \( g(x) = x - 2 \)

1. Find \( f(g(5)) \):
- First, compute \( g(5) \):
\[
g(x) = x - 2 \implies g(5) = 5 - 2 = 3
\]
- Next, substitute \( g(5) = 3 \) into \( f(x) \):
\[
f(x) = 4x + 3 \implies f(g(5)) = f(3) = 4(3) + 3 = 12 + 3 = 15
\]
- Answer: \( f(g(5)) = 15 \)

2. Find \( g(f(-6)) \):
- First, compute \( f(-6) \):
\[
f(x) = 4x + 3 \implies f(-6) = 4(-6) + 3 = -24 + 3 = -21
\]
- Next, substitute \( f(-6) = -21 \) into \( g(x) \):
\[
g(x) = x - 2 \implies g(f(-6)) = g(-21) = -21 - 2 = -23
\]
- Answer: \( g(f(-6)) = -23 \)

3. Find \( f(f(7)) \):
- First, compute \( f(7) \):
\[
f(x) = 4x + 3 \implies f(7) = 4(7) + 3 = 28 + 3 = 31
\]
- Next, substitute \( f(7) = 31 \) into \( f(x) \):
\[
f(x) = 4x + 3 \implies f(f(7)) = f(31) = 4(31) + 3 = 124 + 3 = 127
\]
- Answer: \( f(f(7)) = 127 \)

4. Find \( g(f(x)) \):
- Start with \( f(x) = 4x + 3 \):
\[
f(x) = 4x + 3
\]
- Substitute \( f(x) \) into \( g(x) \):
\[
g(x) = x - 2 \implies g(f(x)) = g(4x + 3) = (4x + 3) - 2 = 4x + 1
\]
- Answer: \( g(f(x)) = 4x + 1 \)

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#### Part 2: Using \( f(x) = 6x^2 \) and \( g(x) = 14x + 4 \)

5. Find \( (f \circ g)(x) \):
- The notation \( (f \circ g)(x) \) means \( f(g(x)) \).
- First, compute \( g(x) \):
\[
g(x) = 14x + 4
\]
- Substitute \( g(x) \) into \( f(x) \):
\[
f(x) = 6x^2 \implies f(g(x)) = f(14x + 4) = 6(14x + 4)^2
\]
- Expand \( (14x + 4)^2 \):
\[
(14x + 4)^2 = (14x)^2 + 2(14x)(4) + 4^2 = 196x^2 + 112x + 16
\]
- Substitute back:
\[
f(g(x)) = 6(196x^2 + 112x + 16) = 1176x^2 + 672x + 96
\]
- Answer: \( (f \circ g)(x) = 1176x^2 + 672x + 96 \)

6. Find \( (g \circ f)(x) \):
- The notation \( (g \circ f)(x) \) means \( g(f(x)) \).
- First, compute \( f(x) \):
\[
f(x) = 6x^2
\]
- Substitute \( f(x) \) into \( g(x) \):
\[
g(x) = 14x + 4 \implies g(f(x)) = g(6x^2) = 14(6x^2) + 4 = 84x^2 + 4
\]
- Answer: \( (g \circ f)(x) = 84x^2 + 4 \)

7. Are these two answers the same? What does this information tell you about composition?
- From the previous steps:
\[
(f \circ g)(x) = 1176x^2 + 672x + 96
\]
\[
(g \circ f)(x) = 84x^2 + 4
\]
- These two expressions are not the same.
- This tells us that function composition is not commutative, i.e., \( f(g(x)) \neq g(f(x)) \) in general.

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#### Part 3: Using \( f(x) = \lfloor x \rfloor \), \( g(x) = 12x \), and \( h(x) = \frac{6}{x} \)

8. Find \( (f \circ g)(5) \):
- First, compute \( g(5) \):
\[
g(x) = 12x \implies g(5) = 12(5) = 60
\]
- Next, substitute \( g(5) = 60 \) into \( f(x) \):
\[
f(x) = \lfloor x \rfloor \implies f(g(5)) = f(60) = \lfloor 60 \rfloor = 60
\]
- Answer: \( (f \circ g)(5) = 60 \)

9. Find \( (f \circ h)(x) \):
- First, compute \( h(x) \):
\[
h(x) = \frac{6}{x}
\]
- Substitute \( h(x) \) into \( f(x) \):
\[
f(x) = \lfloor x \rfloor \implies f(h(x)) = f\left(\frac{6}{x}\right) = \left\lfloor \frac{6}{x} \right\rfloor
\]
- Answer: \( (f \circ h)(x) = \left\lfloor \frac{6}{x} \right\rfloor \)

10. Find \( (h \circ f)(3) \):
- First, compute \( f(3) \):
\[
f(x) = \lfloor x \rfloor \implies f(3) = \lfloor 3 \rfloor = 3
\]
- Next, substitute \( f(3) = 3 \) into \( h(x) \):
\[
h(x) = \frac{6}{x} \implies h(f(3)) = h(3) = \frac{6}{3} = 2
\]
- Answer: \( (h \circ f)(3) = 2 \)

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Final Answers:

1. \( f(g(5)) = \boxed{15} \)
2. \( g(f(-6)) = \boxed{-23} \)
3. \( f(f(7)) = \boxed{127} \)
4. \( g(f(x)) = \boxed{4x + 1} \)
5. \( (f \circ g)(x) = \boxed{1176x^2 + 672x + 96} \)
6. \( (g \circ f)(x) = \boxed{84x^2 + 4} \)
7. No, they are not the same. Function composition is not commutative.
8. \( (f \circ g)(5) = \boxed{60} \)
9. \( (f \circ h)(x) = \boxed{\left\lfloor \frac{6}{x} \right\rfloor} \)
10. \( (h \circ f)(3) = \boxed{2} \)

Final Boxed Answers:
\[
\boxed{15, -23, 127, 4x + 1, 1176x^2 + 672x + 96, 84x^2 + 4, 60, \left\lfloor \frac{6}{x} \right\rfloor, 2}
\]