
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
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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.
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.
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