Chapter 2: Q. 3 (page 126)

Find the domain of each function and evaluate each function at $x = - 1$ .
role="math" localid="1645959311185" $g (x) = \frac{x + 2}{|x + 2|}$

Short Answer

Expert verified

The domain of the function is $\{x | x \neq - 2\}$

The value at $x = - 1$ is $1$ .

Step by step solution

Step 1. Find the domain of the function.

$g (x) = \frac{x + 2}{|x + 2|}$

For domain, denominator cannot be zero.

So,

$|x + 2| \neq 0 x + 2 \neq 0 x \neq - 2$

Therefore, the domain of the function is $x$ belongs to all real numbers except $- 2$ .

Step 2. Find value of g(x) at x=-1

$g (x) = \frac{x + 2}{|x + 2|} g (- 1) = \frac{- 1 + 2}{|- 1 + 2|} g (- 1) = \frac{1}{|1|} g (- 1) = 1$

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Find the domain of each function and evaluate each function at $x = - 1$ .
role="math" localid="1645959311185" $g (x) = \frac{x + 2}{|x + 2|}$

Short Answer

Step by step solution

Step 1. Find the domain of the function.

Step 2. Find value of g(x) at x=-1

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Find the domain of each function and evaluate each function at x=-1.role="math" localid="1645959311185" g(x)=x+2x+2

Short Answer

Step by step solution

Step 1. Find the domain of the function.

Step 2. Find value of g(x) at x=-1

One App. One Place for Learning.

Most popular questions from this chapter

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Probability and Statistics

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Study anywhere. Anytime. Across all devices.

Find the domain of each function and evaluate each function at $x = - 1$ .
role="math" localid="1645959311185" $g (x) = \frac{x + 2}{|x + 2|}$