Question

$f\left(x\right)$ $=\{\begin{array}{c}2x+1;x\le 0\\ x^2;x>0\end{array}$
Then the values of $f(\frac{1}{2}),f(\frac{-1}{2})$ are

• A. $1,-1$
• B. $0,\frac{1}{4}$
• C. $\frac{1}{4},0$
• D. $-1,1$

Answer 1

The correct option is C $\frac{1}{4},0$

$\Rightarrow$ In function $f\left(\frac{1}{2}\right)$, the $x$ is greater than $0$.

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

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

$\Rightarrow$ In function $f\left(\frac{-1}{2}\right)$, the $x$ is less than 0.

$\therefore$ $f\left(x\right)=2x+1$

$\Rightarrow$ $f\left(\frac{-1}{2}\right)=2\times \left(\frac{-1}{2}\right)+1=0$

$\therefore$ The values are $\frac{1}{4}$ and $0$.