Question

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

Answer 1

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 \[2\]
Answer is (a)

Note: A function is said to be differentiable at a point if the left-hand derivative of the function at that point is equal to the right-hand derivative of the function at that point.