# 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

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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 \[2\]

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.

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.

Last updated date: 15th Sep 2023

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