Question

Draw a graph of the function $f\left(x\right)=x-|x-x^2|,-1\le x\le 1$ and discuss its continuity or discontinuity in the interval $-1\le x\le 1.$

Answer 1

Given $f\left(x\right)=x-|x-x^2|$ ; $-1\le x\le 1$

$f\left(x\right)=\{\begin{array}{c}x-(x-x^2)-1\le x\le 0\\ x-[-(x-x^2\left)\right]0\le x\le 1\end{array}$

$f\left(x\right)=\{\begin{array}{c}{x}^2-1\le x\le 0\\ -x^2+2x0\le x\le 1\end{array}$

$f\left(x\right)$ is a polynomial function, which is continuous everywhere. So, is continuous in [-1,0) and also, continuous in (0,1].

We will check the continuity at $x=0$

$LHL={lim}_{x\to 0^{-}}f\left(x\right)$

$={lim}_{x\to 0^{-}}{x}^2$

$=0$

$LHL=0$

$RHL={lim}_{x\to 0^{+}}f\left(x\right)$

$={lim}_{x\to 0^{+}}-x^2+2x$

$=0$

$RHL=0$

Also, $f\left(0\right)=0$

Here, $LHL=RHL=f\left(0\right)$

Hence, $f\left(x\right)$ is continuous at $x=0$

Hence, $f\left(x\right)$ is continuous in [-1,1]