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