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,

& x+a,x\ge -a \\\ & -(x+a),x\le -a \\\ \end{aligned} \right.$$ So , $$\left| x+1 \right|=\left\\{ \begin{aligned} & x+1,\text{ }x\ge -1 \\\ & -\left( x+1 \right),\text{ }x\le -1 \\\ \end{aligned} \right.$$ and $$\left| x-1 \right|=\left\\{ \begin{aligned} & x-1,\text{ }x\ge 1 \\\ & -\left( x-1 \right),\text{ }x\le 1 \\\ \end{aligned} \right.$$ So, we can rewrite the function as $$f\left( x \right)=\left\\{ \begin{aligned} & -(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{aligned} \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.