Question

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

Short answer

The value of function are

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

Step-by-step solution

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

From the given function,

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

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

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

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" $$\begin{gathered} f\left(2\right)=2|\left(2\right)-1|\mathrm{....}\left(\text{Substitute}x=2\text{in}\left(1\right)\right) \\ =2 \\ f\left(x+1\right)=2|\left(x+1\right)-1|\mathrm{....}\left(\text{Substitute}x=x+1\text{\hspace{0.17em}in}\left(1\right)\right) \\ =2\left|x\right| \\ f\left(x^2+2\right)=2\left|\left(x^2+2\right)-1\right|\mathrm{....}\left(\text{Substitute}x=x^2\text{+2in}\left(1\right)\right) \\ =2\left|x^2+1\right| \end{gathered}$$

Step 3. Description of steps.

From the obtained values,

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