Suggested languages for you:

Americas

Europe

Q29.

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 $f\left(-2\right),f\left(0\right),f\left(\frac{1}{2}\right),f\left(2\right),f\left(x+1\right),f\left({x}^{2}+2\right)$ for the function $f\left(x\right)=2\left|x-1\right|$.

The value of function are

$f\left(-2\right)=6,f\left(0\right)=2,f\left(\frac{1}{2}\right)=1,f\left(2\right)=2,f\left(x+1\right)=2\left|x\right|,f\left({x}^{2}+2\right)=2|{x}^{2}+1|$.

See the step by step solution

## Step 1. Evaluate f−2,f0,f12 for the given function.

From the given function,

$f\left(x\right)=2|x-1|....\left(1\right)$

Evaluate the value of function at $x=-2,0,\frac{1}{2}$ gives,

role="math" localid="1644516780178" $f\left(-2\right)=2|\left(-2\right)-1|....\left(\text{Substitute}x=-2\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=2|-3|\phantom{\rule{0ex}{0ex}}=6\phantom{\rule{0ex}{0ex}}f\left(0\right)=2|\left(0\right)-1|....\left(\text{Substitute}x=0\text{\hspace{0.17em}in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=2....\left(\because |-1|=1\right)\phantom{\rule{0ex}{0ex}}f\left(\frac{1}{2}\right)=2\left|\left(\frac{1}{2}\right)-1\right|....\left(\text{Substitute}x=\frac{1}{2}\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=2\left|-\frac{1}{2}\right|\phantom{\rule{0ex}{0ex}}=1$

## Step 2. Evaluate f2,fx+1,fx2+2 for the given function.

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

role="math" localid="1644517286232" $f\left(2\right)=2|\left(2\right)-1|....\left(\text{Substitute}x=2\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}f\left(x+1\right)=2|\left(x+1\right)-1|....\left(\text{Substitute}x=x+1\text{\hspace{0.17em}in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=2\left|x\right|\phantom{\rule{0ex}{0ex}}f\left({x}^{2}+2\right)=2\left|\left({x}^{2}+2\right)-1\right|....\left(\text{Substitute}x={x}^{2}\text{+2 in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=2\left|{x}^{2}+1\right|\phantom{\rule{0ex}{0ex}}$

## Step 3. Description of steps.

From the obtained values,

$f\left(-2\right)=6,f\left(0\right)=2,f\left(\frac{1}{2}\right)=1,f\left(2\right)=2,f\left(x+1\right)=2\left|x\right|,f\left({x}^{2}+2\right)=2|{x}^{2}+1|$.