Question

Let $h\left(x\right)=min\{x,x^2\}$ for every real number of $x$, then which one of the following is true?

• A. $h$ is not continuous for all $x$
• B. $h$ is differentiable for all $x$
• C. $h^{\prime }\left(x\right)=1,$ for all $x$
• D. $h$ is not differentiable at two values of $x$

Answer 1

Answer: (D)

$h$ is not differentiable at two values of $x$

Here, $h\left(x\right)=min\{x,x^2\}$ can be drawn on graph in two steps.
$\left(a\right)$ Draw the graph of $y=x$ and $y=x^2,$ also find their point of intersection.
ie, $x=x^2\Rightarrow x=0,1$
$\left(b\right)$ To find $h\left(x\right)=min\{x;x^2\}$ neglecting the graph above the point of intersection, we get
$h\left(x\right)=x,x\le 0$ or $x\ge 1$
$h\left(x\right)=x^2,0\le x\le 1$
Which shows $h\left(x\right)$ is continuous for all $x$, but not differentiable at $x=0,1$.
Thus, $h\left(x\right)$ is not differentiable at two values of $x$.
Hence, $\left(d\right)$ is the correct answer.