Compose functions worksheet with six problems requiring students to find function compositions and products.

Let's solve each problem step by step. We'll cover function composition and function multiplication.

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🔷 Key Concepts:

- Function Composition:
$(f \circ g)(x) = f(g(x))$ → Plug $g(x)$ into $f$.
$(g \circ f)(x) = g(f(x))$ → Plug $f(x)$ into $g$.

- Function Multiplication:
$(f \cdot g)(x) = f(x) \cdot g(x)$ → Multiply the two functions.

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## ✔ Problem 1:
Given:
$f(n) = 2n - 2$
$g(n) = 5n - 1$

(a) $(f \circ g)(n) = f(g(n))$

$$
g(n) = 5n - 1 \\
f(g(n)) = f(5n - 1) = 2(5n - 1) - 2 = 10n - 2 - 2 = 10n - 4
$$
✔ $(f \circ g)(n) = 10n - 4$

(b) $(g \circ f)(n) = g(f(n))$

$$
f(n) = 2n - 2 \\
g(f(n)) = g(2n - 2) = 5(2n - 2) - 1 = 10n - 10 - 1 = 10n - 11
$$
✔ $(g \circ f)(n) = 10n - 11$

(c) $(f \cdot g)(n) = f(n) \cdot g(n)$

$$
(2n - 2)(5n - 1) = 2n(5n - 1) - 2(5n - 1) = 10n^2 - 2n - 10n + 2 = 10n^2 - 12n + 2
$$
✔ $(f \cdot g)(n) = 10n^2 - 12n + 2$

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## ✔ Problem 2:
Given:
$f(a) = 5a - 6$
$g(a) = 6a - 3$

(a) $(f \circ g)(a) = f(g(a))$

$$
g(a) = 6a - 3 \\
f(g(a)) = f(6a - 3) = 5(6a - 3) - 6 = 30a - 15 - 6 = 30a - 21
$$
✔ $(f \circ g)(a) = 30a - 21$

(b) $(g \circ f)(a) = g(f(a))$

$$
f(a) = 5a - 6 \\
g(f(a)) = g(5a - 6) = 6(5a - 6) - 3 = 30a - 36 - 3 = 30a - 39
$$
✔ $(g \circ f)(a) = 30a - 39$

(c) $(f \cdot g)(a) = f(a) \cdot g(a)$

$$
(5a - 6)(6a - 3) = 5a(6a - 3) - 6(6a - 3) = 30a^2 - 15a - 36a + 18 = 30a^2 - 51a + 18
$$
✔ $(f \cdot g)(a) = 30a^2 - 51a + 18$

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## ✔ Problem 3:
Given:
$f(k) = 8k + 6$
$g(k) = 3k - 6$

(a) $(f \circ g)(k) = f(g(k))$

$$
g(k) = 3k - 6 \\
f(g(k)) = f(3k - 6) = 8(3k - 6) + 6 = 24k - 48 + 6 = 24k - 42
$$
✔ $(f \circ g)(k) = 24k - 42$

(b) $(g \circ f)(k) = g(f(k))$

$$
f(k) = 8k + 6 \\
g(f(k)) = g(8k + 6) = 3(8k + 6) - 6 = 24k + 18 - 6 = 24k + 12
$$
✔ $(g \circ f)(k) = 24k + 12$

(c) $(f \cdot g)(k) = f(k) \cdot g(k)$

$$
(8k + 6)(3k - 6) = 8k(3k - 6) + 6(3k - 6) = 24k^2 - 48k + 18k - 36 = 24k^2 - 30k - 36
$$
✔ $(f \cdot g)(k) = 24k^2 - 30k - 36$

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## ✔ Problem 4:
Given:
$f(c) = 4c - 3$
$g(c) = 2c - 4$

Find:
$(f \circ g)(3)$ and $(g \circ f)(3)$

(a) $(f \circ g)(3) = f(g(3))$

First, find $g(3)$:
$$
g(3) = 2(3) - 4 = 6 - 4 = 2 \\
f(2) = 4(2) - 3 = 8 - 3 = 5
$$
✔ $(f \circ g)(3) = 5$

(b) $(g \circ f)(3) = g(f(3))$

First, find $f(3)$:
$$
f(3) = 4(3) - 3 = 12 - 3 = 9 \\
g(9) = 2(9) - 4 = 18 - 4 = 14
$$
✔ $(g \circ f)(3) = 14$

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## ✔ Problem 5:
Given:
$f(y) = 4y + 5$
$g(y) = 3y + 8$

Find:
$(f \circ g)(-3)$ and $(g \circ f)(-3)$

(a) $(f \circ g)(-3) = f(g(-3))$

First, $g(-3) = 3(-3) + 8 = -9 + 8 = -1$
Then, $f(-1) = 4(-1) + 5 = -4 + 5 = 1$
✔ $(f \circ g)(-3) = 1$

(b) $(g \circ f)(-3) = g(f(-3))$

First, $f(-3) = 4(-3) + 5 = -12 + 5 = -7$
Then, $g(-7) = 3(-7) + 8 = -21 + 8 = -13$
✔ $(g \circ f)(-3) = -13$

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## ✔ Problem 6:
Given:
$f(a) = 8a - 5$
$g(a) = 5a - 2$

Find:
$(f \circ g)(-8)$ and $(g \circ f)(-8)$

(a) $(f \circ g)(-8) = f(g(-8))$

First, $g(-8) = 5(-8) - 2 = -40 - 2 = -42$
Then, $f(-42) = 8(-42) - 5 = -336 - 5 = -341$
✔ $(f \circ g)(-8) = -341$

(b) $(g \circ f)(-8) = g(f(-8))$

First, $f(-8) = 8(-8) - 5 = -64 - 5 = -69$
Then, $g(-69) = 5(-69) - 2 = -345 - 2 = -347$
✔ $(g \circ f)(-8) = -347$

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $(f \circ g)(n) = 10n - 4$, $(g \circ f)(n) = 10n - 11$, $(f \cdot g)(n) = 10n^2 - 12n + 2$ |
| 2 | $(f \circ g)(a) = 30a - 21$, $(g \circ f)(a) = 30a - 39$, $(f \cdot g)(a) = 30a^2 - 51a + 18$ |
| 3 | $(f \circ g)(k) = 24k - 42$, $(g \circ f)(k) = 24k + 12$, $(f \cdot g)(k) = 24k^2 - 30k - 36$ |
| 4 | $(f \circ g)(3) = 5$, $(g \circ f)(3) = 14$ |
| 5 | $(f \circ g)(-3) = 1$, $(g \circ f)(-3) = -13$ |
| 6 | $(f \circ g)(-8) = -341$, $(g \circ f)(-8) = -347$ |

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