Question

Rolle’s theorem is not applicable to the function $f(x) = |x|$ defined on $\left[ { - 1,1} \right]$ because
A. F is not continuous on $\left[ { - 1,1} \right]$
B. F is not differentiable on $\left( { - 1,1} \right)$
C. $f\left( { - 1} \right) \ne f\left( 1 \right)$
D. $f\left( { - 1} \right) = f\left( 1 \right) \ne 0$

Answer 1

Hint: Given, $f(x) = |x|$ defined on $\left[ { - 1,1} \right]$. We have to find why Rolle’s theorem is not applicable to the function given $f(x)$. First, we will properly define $f(x)$. Then we will check that $f(x)$ is continuous or not. If $f(x)$ is continuous then we will check differentiability.

Formula Used: $Lf’(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{h}$
$Rf’(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$

Complete step by step solution: According to Rolle's theorem, a real-valued differentiable function must have at least one fixed point where the first derivative is zero if it achieves equal values at two different places. Rolle's theorem is named after A French mathematician named Michel Rolle. The mean value theorem has a specific case known as Rolle's Theorem.
There must be at least one tangent to the curve that is parallel to the x-axis if the curve $y = f(x)$ is continuous between $x = a$ and $x = b$, a tangent can be drawn at any point within the interval, and the ordinates corresponding to the abscissa and are the same.
According to this theorem, if f $f(x)$ denotes a polynomial function in $x$ and $x = a$ and $b$ are the two roots of the equation \[f\left( x \right){\text{ }} = {\text{ }}0\], then at least one root of the equation \[f'\left( x \right){\text{ }} = {\text{ }}0\] must be in the middle of these two numbers.
Given, $f(x) = |x|$
We know that
$|x| = - x,\,\,if\,x < 0$
And $|x| = x,\,\,if\,x > 0$
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  { - x}&{if\,x < 0} \\
  x&{if\,x > 0}
\end{array}} \right.$
Now, $f( - 1) = | - 1|$
$ = 1$
$f(1) = |1|$
$ = 1$
$ \Rightarrow f( - 1) = 1 = f(1)$
Now, we will find the LHD and RHD
$Lf’(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \dfrac{{| - h|}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{ - h}}$
$ = - 1$
$Rf’(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{|h|}}{h}\]
\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{h}\]
$ = 1$
Clearly $LHD \ne RHD$
Hence $f(x)$ is not differentiable on $\left( { - 1,1} \right)$
Therefore, option B is correct.

Note: For solving this question students must have knowledge about Rolle’s theorem. They should know how to check whether a function is continuous or not. After checking continuity we will check whether the function is differentiable. And should check differentiability by checking if LHD is equal to RHD or not.