Question

Let $h\left(x\right)=min\{x,x^2\}$ for $x\in R$. Then which of the following is correct

• A. $h$ is continuous for all $x$
• B. $h$ is differentiable for all $x$
• C. $h^{\prime }\left(x\right)=1$ for all $x>1$
• D. $h$ is not differentiable at $2$ points

Answer 1

Answer: (A), (C), (D)

$h$ is continuous for all $x$
$h^{\prime }\left(x\right)=1$ for all $x>1$
$h$ is not differentiable at $2$ points

$h\left(x\right)=\{\begin{array}{c}x;x\ge 1\\ x^2;0\le x<1\\ x;x<0\end{array}$

From the graph it is clear that $h$ is continuous.

Also $h$ is differentiable at all $x$ except $x=0$ and $x=1$

$h^{\prime }\left(x\right)=\{\begin{array}{c}1;x>1\\ 2x;0<x<1\\ 1;x<0\end{array}$

For $x=1,h^{\prime }\left(1^{+}\right)={lim}_{t\to 0^{+}}\frac{h(1+t)-h\left(1\right)}{t}={lim}_{t\to 0^{+}}\frac{1+t-1}{t}=1$

But $h^{\prime }\left(1^{-}\right)={lim}_{t\to 0^{+}}\frac{h(1-t)-h\left(1\right)}{t}={lim}_{t\to 0^{+}}\frac{1-t-1}{t}=-2$

So $h$ is not differentiable at $x=1$

Similarly, $h^{\prime }\left(0^{+}\right)=0$ but $h^{\prime }\left(0^{-}\right)=1$

So $h$ is not differentiable at $x=0$ also.

Hence, option C and D.