Question

$f\left(x\right)$ $=\{\begin{array}{c}{x}^2-1;x\le -2\\ x^2;-2<x<2\\ x^2+1;x\ge 2\end{array}$
The value of $f(\frac{1}{2})$ is

• A. $-2$
• B. $2$
• C. $\frac{-1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

$\Rightarrow$ In function $f\left(\frac{1}{2}\right)$, $x$ is greater than $-2$ and less than $2$.

$\Rightarrow$ $f\left(x\right)=x^2$

$\Rightarrow$ $f\left(\frac{1}{2}\right)=(\frac{1}{2}{)}^2$

$\therefore$ $f\left(\frac{1}{2}\right)=\frac{1}{4}$