Question

If a function $f\left(x\right)$ is defined as $f\left(x\right)=\{\begin{array}{cc}-x,& x<0\\ x^2,& 0\le x\le 1\\ x^2-x+1,& x>1\end{array},$
then which among the following option is correct

• A. $f\left(x\right)$ is differentiable function
• B. $f\left(x\right)$ is not differentiable at $x=1$
• C. $f\left(x\right)$ is not differentiable at $x=0$
• D. $f\left(x\right)$ is not differentiable at $x=0,1$

Answer 1

The correct option is D $f\left(x\right)$ is not differentiable at $x=0,1$

$f\left(x\right)$ is defined as $f\left(x\right)=\{\begin{array}{cc}-x,& x<0\\ x^2,& 0\le x\le 1\\ x^2-x+1,& x>1\end{array}$

$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right),\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$
So, $f\left(x\right)$ is continuous function

Now, $f^{\prime }\left(x\right)=\{\begin{array}{cc}-1,& x<0\\ 2x,& 0<x<1\\ 2x-1,& x>1\end{array}$

At $x=0,$ $f^{\prime }\left(0^{-}\right)=-1,f^{\prime }\left(0^{+}\right)=2\left(0\right)=0$
At $x=1,$$f^{\prime }\left(1^{-}\right)=2\left(1\right)=2,f^{\prime }\left(1^{+}\right)=2\left(1\right)-1=1$
Hence we find that $f^{\prime }\left(0^{-}\right)\ne f^{\prime }\left(0^{+}\right),f^{\prime }\left(1^{-}\right)\ne f^{\prime }\left(1^{+}\right)$
$\therefore f\left(x\right)$ is not differentiable at $x=0,1.$