Let $\mathrm{f}(x)=\min \left\{x, x^2\right\}$ for every real number of $x$. then
Let $\mathrm{f}(x)=\min \left\{x, x^2\right\}$ for every real number of $x$. then
$f(x)$ is continuous for all $x$
$f(x)$ is differentiable for all $x$
$f^{\prime}(x)=2$ for all $x\gt1$
$f(x)$ is not differentiable at three values of $x$
Solution
$\begin{aligned} f(x) & =\min \left\{x, x^2\right\} \\ & =\left\{\begin{array}{ccc}x & \text { if } & x \lt 0 \\ x^2 & \text { if } & 0 \leq x \lt 1 \\ x & \text { if } & 1 \lt x\end{array}\right.\end{aligned}$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0$
Similarly, we can show $\lim _{x \rightarrow 0^{+}} f(x)=0$
So, $f(x)$ is continuous at $x=0$
Similarly $f(x)$ is also continuous at $x=1$
and $f^{\prime}(x)=\left\{\begin{array}{cc}1, & x \lt 0 \\ 2 x, & 0 \leq x \lt 1 \\ 1, & x\gt1\end{array}\right.$
So, $f(x)$ is not differentiable at $x=0,1$.