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# The number of points at which the function $f\left( x \right)=\left| x+1 \right|+\left| x-1 \right|$ is not differentiable is:(a) $2$ (b) $1$(c) $0$(d) infinite

Last updated date: 23rd Mar 2023
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Hint: A function is differentiable at $x=a$ , if the left-hand derivative of the function is equal to the right hand derivative of the function at $x=a$.

The given function is $f\left( x \right)=\left| x+1 \right|+\left| x-1 \right|$
We know,
\left| x+a \right|=\left\{ \begin{align} & x+a,x\ge -a \\ & -(x+a),x\le -a \\ \end{align} \right.
So , \left| x+1 \right|=\left\{ \begin{align} & x+1,\text{ }x\ge -1 \\ & -\left( x+1 \right),\text{ }x\le -1 \\ \end{align} \right.
and \left| x-1 \right|=\left\{ \begin{align} & x-1,\text{ }x\ge 1 \\ & -\left( x-1 \right),\text{ }x\le 1 \\ \end{align} \right.
So, we can rewrite the function as
f\left( x \right)=\left\{ \begin{align} & -(x+1)-\left( x-1 \right),\text{ }x\le -1 \\ & \left( x+1 \right)-\left( x-1 \right),\text{ -1}\le \text{ }x\le 1 \\ & \left( x+1 \right)+\left( x-1 \right),\text{ }x\ge 1 \\ \end{align} \right.
Now, we can see there are two critical points i.e. $x=-1$and $x=1$.
We will check if the function is differentiable at critical points of the function .
A function is differentiable at $x=a$ , if the left-hand derivative of the function is equal to the right hand derivative of the function at $x=a$.
We know , the left hand derivative of $f\left( x \right)$at $x=a$is given as ${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}$ and the right hand derivative of $f\left( x \right)$at $x=a$is given as ${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$.

Now , we will check the differentiability of the function at $x=-1$.
The right-hand derivative of the function at $x=-1$is given by
${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( -1+h \right)-f\left( -1 \right)}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( -1+h+1 \right)-\left( -1+h-1 \right)-\left( 2 \right)}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{h-\left( -2+h \right)-\left( 2 \right)}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{h+2-h-2}{h}=0$
Now , the left-hand derivative of $f\left( x \right)$at $x=-1$is given as
${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( -1-h \right)-f\left( -1 \right)}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\left( -1-h+1 \right)-\left( -1-h-1 \right)-2}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{h+2+h-2}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2h}{-h}=-2$
Clearly , the left-hand derivative of the function at $x=-1$ is not equal to the right hand derivative of the function at $x=-1$.
So , the function is not differentiable at $x=-1$.
Now, we will check the differentiability of $f\left( x \right)$ at $x=1$
The left-hand derivative of $f\left( x \right)$at $x=1$is given by
${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1-h+1 \right)-\left( 1-h-1 \right)-2}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2-h+h-2}{-h}=0$
Now , the right-hand derivative of $f\left( x \right)$at $x=1$ is given by
${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+h+1 \right)+\left( 1+h-1 \right)-\left( 2 \right)}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2+h+h-2}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2h}{h}=2$
Clearly, ${{L}^{'}}\ne {{R}^{'}}$, i.e. left-hand derivative of the function at $x=1$ is not equal to right hand derivative of the function at $x=1$. So , the function $f\left( x \right)$is not differentiable at $x=1$.
So , the number of points at which the function is not differentiable is $2$