Question

The number of points at which $f\left(x\right)=\{\begin{array}{cc}min(x,x^2),& \text{if}-\infty <x<1\\ min(2x-1,x^2),& \text{if}x\ge 1\end{array}$ is not differentiable is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$f\left(x\right)=\{\begin{array}{cc}min(x,x^2),& \text{if}-\infty <x<1\\ min(2x-1,x^2),& \text{if}x\ge 1\end{array}$




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

As $f^{\prime }\left(0^{-}\right)\ne f^{\prime }\left(0^{+}\right)$, so non differentiable at $x=0$
But $f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$, so differentiable at $x=1$

Hence, $f\left(x\right)$ is non differentiable at $1$ point.