Question

Let $h\left(\:x\:\right)\:=\:\min \:\left\{\:x,\:x^2\:\right\}$, for every real number $x$. Then
A. $h$ is discontinuous for all $x$
B. $h$ is differentiable for all $x$
C. $h'\left( x \right) = 2$ for all $x > 1$
D. $h$ is not differentiable at two values of $x$

Answer 1

Hint: Write the function as a multivalued function. Then check the continuity and differentiability at the breaking points.

Formula Used:
A function $f\left( x \right)$ is continuous at the point $x = a$ if and only if $ {\lim }_{x \to {a^ - }} f\left( x \right) = {\lim }_{x \to {a^ + }} f\left( x \right) = f\left( a \right)$
A function $f\left( x \right)$ is differentiable at the point $x = a$ if and only if $Lf'\left( a \right) = Rf'\left( a \right)$, where $Lf'\left( a \right) = {\lim }_{x \to {a^ - }} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}$ and $Rf'\left( a \right) = {\lim }_{x \to {a^ + }} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}$, provided the limit exists.

Complete step by step solution:
The given function is $h\left(\:x\:\right)\:=\:\min \:\left\{\:x,\:x^2\:\right\}$
When $x\:<\:0,\:\min \:\left\{\:x,\:x^2\:\:\right\}=\:x$ because $x$ takes only negative values but ${x^2}$ takes all positive values.
When $x\:=\:0,\:\min \:\left\{\:x,\:x^2\:\:\right\}=\:0$ because $x = {x^2} = 0$
When $0 < x < 1$, $\min \:\left\{\:x,\:x^2\:\:\right\}=\:x^2$ because the series of proper fractions is decreasing.
When $x\:=\:1,\:\min \:\left\{\:x,\:x^2\:\right\} = 1$ because $x = {x^2} = 1$
When $x\:>\:1,\:\min \:\left\{\:x,\:x^2\:\right\} = x$ because ${x^2} > x$
So, $h\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{x^2},}&{0 < x < 1}\\{x,}&{otherwise}\end{array}} \right.$
At $x = 0$ and $x = 1$, the values of $h\left( x \right)$ are $0$ and $1$ respectively.
Check the continuity of $h\left( x \right)$ at $x = 0$ and $x = 1$
At $x = 0$:-
$ {\lim }_{x \to {0^ - }} h\left( x \right) = {\lim }_{x \to {0^ - }} \left( x \right) = 0$
$ {\lim }_{x \to {0^ + }} h\left( x \right) = {\lim }_{x \to {0^ + }} \left( {{x^2}} \right) = 0$
and $h\left( 0 \right) = 0$
$\therefore {\lim }_{x \to {0^ - }} h\left( x \right) = {\lim }_{x \to {0^ + }} h\left( x \right) = h\left( 0 \right)$
So, $h\left( x \right)$ is continuous at $x = 0$
At $x = 1$:-
$ {\lim }_{x \to {1^ - }} h\left( x \right) = {\lim }_{x \to {1^ - }} \left( {{x^2}} \right) = 1$
$ {\lim }_{x \to {1^ + }} h\left( x \right) = {\lim }_{x \to {1^ + }} \left( x \right) = 1$
and $h\left( 1 \right) = 1$
$\therefore {\lim }_{x \to {1^ - }} h\left( x \right) = {\lim }_{x \to {1^ + }} h\left( x \right) = h\left( 1 \right)$
So, $h\left( x \right)$ is continuous at $x = 1$
Check the differentiability of $h\left( x \right)$ at $x = 0$ and $x = 1$
$\therefore Lh'\left( 0 \right) \ne Rh'\left( 0 \right)$
So, $h\left( x \right)$ is not differentiable at $x = 0$
$\therefore Lh'\left( 1 \right) \ne Rh'\left( 1 \right)$
So, $h\left( x \right)$ is not differentiable at $x = 1$

Option ‘D’ is correct

Note: Every polynomial function is continuous and differentiable everywhere. So, $h\left( x \right)$ is obviously continuous and differentiable in the intervals $\left( {0,1} \right)$ and $\mathbb{R}\:-\:\left\{\:0,1\:\right\}$. The function changes the form at $x = 0$ and $x = 1$. So, you must check its continuity and differentiability at these two points.