Composition of Functions Worksheets - Free Printable

Let's solve each problem step by step from the worksheet titled "Evaluating Composition of Two Functions."

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Problem 1:

Given:
- $ f(x) = x + 3 $
- $ g(x) = x^2 $

Find:
(a) $ (f \circ g)(x) $
(b) $ (g \circ f)(x) $

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

First, compute $ g(x) = x^2 $, then plug it into $ f $:

$$
f(g(x)) = f(x^2) = x^2 + 3
$$

✔ So, $ (f \circ g)(x) = x^2 + 3 $

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#### (b) $ (g \circ f)(x) = g(f(x)) $

First, compute $ f(x) = x + 3 $, then plug it into $ g $:

$$
g(f(x)) = g(x + 3) = (x + 3)^2 = x^2 + 6x + 9
$$

✔ So, $ (g \circ f)(x) = x^2 + 6x + 9 $

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✔ Answer 1:

(a) $ x^2 + 3 $
(b) $ x^2 + 6x + 9 $

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Problem 2:

Given:
- $ f(x) = 2x - 7 $
- $ g(x) = x^2 - 4 $

Evaluate:
(a) $ (f \circ g)(-2) $
(b) $ (g \circ f)(-2) $

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#### (a) $ (f \circ g)(-2) = f(g(-2)) $

Step 1: Compute $ g(-2) $
$$
g(-2) = (-2)^2 - 4 = 4 - 4 = 0
$$

Step 2: Plug into $ f $:
$$
f(0) = 2(0) - 7 = -7
$$

✔ So, $ (f \circ g)(-2) = -7 $

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#### (b) $ (g \circ f)(-2) = g(f(-2)) $

Step 1: Compute $ f(-2) $
$$
f(-2) = 2(-2) - 7 = -4 - 7 = -11
$$

Step 2: Plug into $ g $:
$$
g(-11) = (-11)^2 - 4 = 121 - 4 = 117
$$

✔ So, $ (g \circ f)(-2) = 117 $

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✔ Answer 2:

(a) $ -7 $
(b) $ 117 $

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Problem 3:

Given:
- $ f(x) = \sqrt{2x + 1} $
- $ g(x) = 3x^2 + 1 $

Evaluate:
(a) $ (f \circ g)(2) $
(b) $ (g \circ f)\left(\frac{4}{5}\right) $

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#### (a) $ (f \circ g)(2) = f(g(2)) $

Step 1: Compute $ g(2) $
$$
g(2) = 3(2)^2 + 1 = 3(4) + 1 = 12 + 1 = 13
$$

Step 2: Plug into $ f $:
$$
f(13) = \sqrt{2(13) + 1} = \sqrt{26 + 1} = \sqrt{27} = 3\sqrt{3}
$$

✔ So, $ (f \circ g)(2) = 3\sqrt{3} $

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#### (b) $ (g \circ f)\left(\frac{4}{5}\right) = g\left(f\left(\frac{4}{5}\right)\right) $

Step 1: Compute $ f\left(\frac{4}{5}\right) $
$$
f\left(\frac{4}{5}\right) = \sqrt{2\left(\frac{4}{5}\right) + 1} = \sqrt{\frac{8}{5} + 1} = \sqrt{\frac{8}{5} + \frac{5}{5}} = \sqrt{\frac{13}{5}}
$$

Step 2: Plug into $ g $:
$$
g\left(\sqrt{\frac{13}{5}}\right) = 3\left(\sqrt{\frac{13}{5}}\right)^2 + 1 = 3\left(\frac{13}{5}\right) + 1 = \frac{39}{5} + 1 = \frac{39}{5} + \frac{5}{5} = \frac{44}{5}
$$

✔ So, $ (g \circ f)\left(\frac{4}{5}\right) = \frac{44}{5} $

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✔ Answer 3:

(a) $ 3\sqrt{3} $
(b) $ \frac{44}{5} $

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Problem 4:

Given:
- $ h(x) = \frac{4}{x} $
- $ k(x) = \frac{1}{x - 1} $

Find $ (h \circ k)\left(\frac{1}{2}\right) $

That is:
$$
(h \circ k)\left(\frac{1}{2}\right) = h\left(k\left(\frac{1}{2}\right)\right)
$$

Step 1: Compute $ k\left(\frac{1}{2}\right) $
$$
k\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2} - 1} = \frac{1}{-\frac{1}{2}} = -2
$$

Step 2: Plug into $ h $:
$$
h(-2) = \frac{4}{-2} = -2
$$

✔ So, $ (h \circ k)\left(\frac{1}{2}\right) = -2 $

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✔ Answer 4:

$ -2 $

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Problem 5:

Given:
- $ f(x) = x^2 $ and $ g(x) = \frac{x - 1}{x} $

Which of the following represents $ (g \circ f)(x) $?
Options:
A. $ \frac{x^2 - 1}{x} $
B. $ \frac{x^2 - 1}{x^2} $
C. $ \frac{x^2}{x - 1} $
D. $ \frac{x^2 - 1}{x - 1} $

We need to find $ (g \circ f)(x) = g(f(x)) $

Step 1: $ f(x) = x^2 $

Step 2: $ g(f(x)) = g(x^2) = \frac{x^2 - 1}{x^2} $

✔ So, $ (g \circ f)(x) = \frac{x^2 - 1}{x^2} $

This matches Option B.

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✔ Answer 5:

B. $ \frac{x^2 - 1}{x^2} $

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Problem 6:

Given:
- $ p(x) = 2x $ and $ q(x) = \frac{1}{x^2} $

What is $ (p \circ q)\left(\frac{1}{2}\right) $?

So:
$$
(p \circ q)\left(\frac{1}{2}\right) = p\left(q\left(\frac{1}{2}\right)\right)
$$

Step 1: $ q\left(\frac{1}{2}\right) = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4 $

Step 2: $ p(4) = 2(4) = 8 $

✔ So, $ (p \circ q)\left(\frac{1}{2}\right) = 8 $

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✔ Answer 6:

$ 8 $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1a | $ x^2 + 3 $ |
| 1b | $ x^2 + 6x + 9 $ |
| 2a | $ -7 $ |
| 2b | $ 117 $ |
| 3a | $ 3\sqrt{3} $ |
| 3b | $ \frac{44}{5} $ |
| 4 | $ -2 $ |
| 5 | B |
| 6 | $ 8 $ |

Let me know if you'd like these written in a formatted worksheet style!