Question

Let $h\left(x\right)=$min{$x$, $x^2$} for every real number $x$. Then

• 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 two values of $x$

Answer 1

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

The correct options are
A $h$ is continuous for all $x$
C $h^{\prime }\left(x\right)=1$, for all $x>1$
D $h$ is not differentiable at two values of $x$

We first have to draw the graph of both the functions and then to check for $h\left(x\right)$

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

$h^{\prime }\left(x\right)=\{\begin{array}{c}1x<0\\ 2x0<x<1\\ 1x\ge 1\end{array}$

$\Rightarrow h^{\prime }\left(1^{-}\right)=2$ and $h^{\prime }\left(1^{+}\right)=1$

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

Hence,

1. $h\left(x\right)$ is continuous for all $x$
2. $h\left(x\right)$ is not differentiable for all $x$
3. $h^{\prime }\left(x\right)=1$ for $x>1$
4. $h\left(x\right)$ is not differentiable at two values of $x$, i.e. $x=0,x=1$