## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 2 - Section 2.1 - Functions - Exercises: 30

#### Answer

Evaluate $f(x) = \frac{|x|}{x}$: $f(-2) = -1$ $f(-1) = -1$ $f(0) = undefined$ $f(5) = 1$ $f(x^2) = 1$ $f(\frac{1}{x}) = \pm 1$

#### Work Step by Step

For x = -2 $f(-2) = \frac{|-2|}{-2}$ ... take absolute value of numerator $=\frac{2}{-2}$ ... simplify $=-1$ _________________ For x = -1 $f(-1) = \frac{|-1|}{-1}$ ... take absolute value of numerator $=\frac{1}{-1}$ ... simplify $=-1$ _________________ For x = 0 $f(-2) = \frac{|0|}{0}$ ... take absolute value of numerator $=\frac{0}{0} = undefined$ _________________ For x = 5 $f(-2) = \frac{|5|}{5}$ ... take absolute value of numerator $=\frac{5}{5}$ ... simplify $=1$ _________________ For x = $x^2$ $f(-2) = \frac{|x^2|}{x^2}$ ... take absolute value of numerator $=\frac{x^2}{x^2}$ ... simplify $=1$ _________________ For x = $\frac{1}{x}$ $f(\frac{1}{x}) = \frac{|\frac{1}{x}|}{\frac{1}{x}}$ ... take absolute value of numerator $=\frac{\frac{1}{|x|}}{\frac{1}{x}}$ ... simplify $=\frac{x}{|x|} = \pm 1$ _________________

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