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Expert-verified Found in: Page 156 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Find the value of $f\left(-2\right),f\left(-1\right),f\left(0\right),f\left(5\right),f\left({x}^{2}\right),f\left(\frac{1}{x}\right)$ for the function $f\left(x\right)=\frac{\left|x\right|}{x}$.

The value of function are

$f\left(-2\right)=-1,\text{ }f\left(-1\right)=-1,f\left(0\right)=\text{undefined},f\left(5\right)=1,f\left({x}^{2}\right)=1,f\left(\frac{1}{x}\right)=1$.

See the step by step solution

## Step 1. Evaluate f−2,f−1,f0 for the given function.

From the given function,

$f\left(x\right)=\frac{|x|}{x}....\left(1\right)$

Evaluate the value of function at $x=-2,-1,0$ gives,

role="math" localid="1644518088722" $f\left(-2\right)=\frac{|-2|}{-2}....\left(\text{Substitute}x=-2\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-1\phantom{\rule{0ex}{0ex}}f\left(-1\right)=\frac{|\left(-1\right)|}{\left(-1\right)}....\left(\text{Substitute}x=-1\text{\hspace{0.17em}in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-1\phantom{\rule{0ex}{0ex}}f\left(0\right)=\frac{\left|\left(0\right)\right|}{\left(0\right)}....\left(\text{Substitute}x=0\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}\left(\text{Which is undefined}\right)$

## Step 2. Evaluate f5,fx2,f1x for the given function.

Similarly, evaluate the value of function (1) at $x=5,{x}^{2},\frac{1}{x}$ gives,

role="math" localid="1644518425397" $f\left(5\right)=\frac{|\left(5\right)|}{\left(5\right)}....\left(\text{Substitute}x=5\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}f\left({x}^{2}\right)=\frac{|\left({x}^{2}\right)|}{\left({x}^{2}\right)}....\left(\text{Substitute}x={x}^{2}\text{\hspace{0.17em}in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}f\left(\frac{1}{x}\right)=\frac{\left|\left(\frac{1}{x}\right)\right|}{\left(\frac{1}{x}\right)}....\left(\text{Substitute}x=\frac{1}{x}\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}\text{=1}$

## Step 3. Description of steps.

From the obtained values,

$f\left(-2\right)=-1,\text{ }f\left(-1\right)=-1,f\left(0\right)=\text{undefined},f\left(5\right)=1,f\left({x}^{2}\right)=1,f\left(\frac{1}{x}\right)=1$. ### Want to see more solutions like these? 