Solved Functions Worksheet Domain Range and Function | Chegg.com - Free Printable

Let's solve each problem step by step.

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1. Find the domain

The domain of a function is the set of all real numbers for which the function is defined (i.e., where the expression makes sense, especially avoiding division by zero or square roots of negative numbers).

#### a. $ f(x) = \frac{x - 4}{x - 2} $

- This is a rational function.
- The denominator cannot be zero: $ x - 2 \neq 0 $
- So, $ x \neq 2 $
- Domain: All real numbers except $ x = 2 $
- Answer: $ (-\infty, 2) \cup (2, \infty) $

#### b. $ g(x) = \frac{x^2 + 5}{x + 1} $

- Denominator: $ x + 1 \neq 0 \Rightarrow x \neq -1 $
- Domain: All real numbers except $ x = -1 $
- Answer: $ (-\infty, -1) \cup (-1, \infty) $

#### c. $ h(x) = \frac{x}{x^2 - 9} $

- Denominator: $ x^2 - 9 = (x - 3)(x + 3) $
- Cannot be zero → $ x \neq 3 $ and $ x \neq -3 $
- Domain: All real numbers except $ x = -3 $ and $ x = 3 $
- Answer: $ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) $

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2. Let $ f(x) = 2x - 1 $ and $ g(x) = x^2 - 4 $. Find:

We'll evaluate each expression using substitution.

#### a. $ f(0) $

$ f(0) = 2(0) - 1 = -1 $

✔ Answer: $ -1 $

#### b. $ f(1) $

$ f(1) = 2(1) - 1 = 1 $

✔ Answer: $ 1 $

#### c. $ f(-1) $

$ f(-1) = 2(-1) - 1 = -2 - 1 = -3 $

✔ Answer: $ -3 $

#### d. $ f(a) $

Replace $ x $ with $ a $:
$ f(a) = 2a - 1 $

✔ Answer: $ 2a - 1 $

#### e. $ g(0) $

$ g(0) = (0)^2 - 4 = -4 $

✔ Answer: $ -4 $

#### f. $ g(-2) $

$ g(-2) = (-2)^2 - 4 = 4 - 4 = 0 $

✔ Answer: $ 0 $

#### g. $ g(3) $

$ g(3) = (3)^2 - 4 = 9 - 4 = 5 $

✔ Answer: $ 5 $

#### h. $ g(t) $

Replace $ x $ with $ t $:
$ g(t) = t^2 - 4 $

✔ Answer: $ t^2 - 4 $

#### i. $ f(a+5) $

Plug $ a+5 $ into $ f(x) $:
$ f(a+5) = 2(a+5) - 1 = 2a + 10 - 1 = 2a + 9 $

✔ Answer: $ 2a + 9 $

#### j. $ f(x+h) $

Plug $ x+h $ into $ f(x) $:
$ f(x+h) = 2(x+h) - 1 = 2x + 2h - 1 $

✔ Answer: $ 2x + 2h - 1 $

#### k. $ g(a-1) $

Plug $ a-1 $ into $ g(x) $:
$ g(a-1) = (a-1)^2 - 4 = a^2 - 2a + 1 - 4 = a^2 - 2a - 3 $

✔ Answer: $ a^2 - 2a - 3 $

#### l. $ g(f(x)) $

This is a composite function: plug $ f(x) $ into $ g(x) $

First, $ f(x) = 2x - 1 $

Now, $ g(f(x)) = g(2x - 1) = (2x - 1)^2 - 4 $

Compute:
$$
(2x - 1)^2 = 4x^2 - 4x + 1
$$
Then:
$$
g(f(x)) = 4x^2 - 4x + 1 - 4 = 4x^2 - 4x - 3
$$

✔ Answer: $ 4x^2 - 4x - 3 $

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3. If $ f(x) = 3x - 5 $, find $ \frac{f(x) - f(a)}{x - a} $

This is the difference quotient, often used in calculus to find slopes.

We are given:
$$
f(x) = 3x - 5, \quad f(a) = 3a - 5
$$

So:
$$
f(x) - f(a) = (3x - 5) - (3a - 5) = 3x - 5 - 3a + 5 = 3x - 3a = 3(x - a)
$$

Now divide by $ x - a $:
$$
\frac{f(x) - f(a)}{x - a} = \frac{3(x - a)}{x - a} = 3 \quad \text{(for } x \ne a\text{)}
$$

✔ Answer: $ 3 $

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✔ Final Answers:

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#### 1. Domain
a. $ (-\infty, 2) \cup (2, \infty) $
b. $ (-\infty, -1) \cup (-1, \infty) $
c. $ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) $

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#### 2. Evaluations
a. $ f(0) = -1 $
b. $ f(1) = 1 $
c. $ f(-1) = -3 $
d. $ f(a) = 2a - 1 $
e. $ g(0) = -4 $
f. $ g(-2) = 0 $
g. $ g(3) = 5 $
h. $ g(t) = t^2 - 4 $
i. $ f(a+5) = 2a + 9 $
j. $ f(x+h) = 2x + 2h - 1 $
k. $ g(a-1) = a^2 - 2a - 3 $
l. $ g(f(x)) = 4x^2 - 4x - 3 $

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#### 3. Difference Quotient
$$
\frac{f(x) - f(a)}{x - a} = 3
$$

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