Question

Let $f\left(x\right)=\{\begin{array}{cc}2+\sqrt{1-x^2},& \left|x\right|\le 1\\ 2e^{{(1-x)}^2},& \left|x\right|>1\end{array}$
The points where $f\left(x\right)$ is not differentiable is/are :

• A. $0$
• B. $-1,1$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $-1,1$

Given function is $f\left(x\right)=\{\begin{array}{ccc}2+\sqrt{1-x^2}& ,& \left|x\right|\le 1\\ 2e^{{(1-x)}^2}& ,& \left|x\right|>1\end{array}$
Now, to check the differentiability of function
$f^{\prime }\left(x\right)=\{\begin{array}{ccc}4e^{{(1-x)}^2}(x-1)& ,& x\in (-\infty ,-1)\\ \frac{-x}{\sqrt{1-x^2}}& ,& x\in (-1,1)\\ 4e^{{(1-x)}^2}(x-1)& ,& x\in (1,\infty )\end{array}$

Clearly, left hand derivative at $x=1$
$f^{\prime }\left(1^{-}\right)$ is not defined.
Also, right hand derivative at $x=-1$
$f^{\prime }(-1^{+})$ is not defined.
Thus, $f\left(x\right)$ is not differentiable at $x=-1,1$.