# Discuss the applicability of Lagrange's mean value theorem for the function $f\left( x \right) = \left| x \right|$ on $\left[ { - 1,1} \right]$.

Last updated date: 24th Mar 2023

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Hint: In this question, for applicability of Lagrange's Mean Value theorem on f(x) in interval $\left[ {a,b} \right]$, function must be continuous in $\left[ {a,b} \right]$ and differentiable in (a,b).

Complete step-by-step answer:

Now we define the function $f\left( x \right) = \left| x \right|$ on $\left[ { - 1,1} \right]$ as follows,

$

f\left( x \right) = - x,x \in \left[ { - 1,0} \right) \\

f\left( x \right) = x,x \in \left[ {0,1} \right] \\

$

Now let's examine continuity and differentiability of function at x=0 .

For continuity,

\[

{\text{Left hand limit,}}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0 \\

{\text{Right hand limit,}}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = 0 \\

\]

Since, LHL=RHL, f(x) is continuous at x=0 and f(0)=0 .

For differentiability,

Left hand derivative

$

f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x - h} \right) - f\left( x \right)}}{{ - h}} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( { - h} \right) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \left( { - h} \right) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{ - h}} = - 1 \\

$

Right hand derivative

$

f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( h \right) - 0}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{h} = 1 \\

$

Since, $LHD \ne RHD$ , f(x) is not differentiable at x=0

For applicability of Lagrange's Mean Value theorem on f(x) in interval $\left[ { - 1,1} \right]$ , Function must be continuous in $\left[ { - 1,1} \right]$ and differentiable in (1,1) .

But the function is not differentiable at x=0 .

So, Lagrange's Mean Value theorem is not applicable for f(x) in interval $\left[ { - 1,1} \right]$.

Note: Whenever we face such types of problems we use some important points. First we define the function on different intervals then check the continuity and differentiability of the function on different intervals. If function be continuous and differentiable on interval then Lagrange's mean value theorem be applicable.

Complete step-by-step answer:

Now we define the function $f\left( x \right) = \left| x \right|$ on $\left[ { - 1,1} \right]$ as follows,

$

f\left( x \right) = - x,x \in \left[ { - 1,0} \right) \\

f\left( x \right) = x,x \in \left[ {0,1} \right] \\

$

Now let's examine continuity and differentiability of function at x=0 .

For continuity,

\[

{\text{Left hand limit,}}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0 \\

{\text{Right hand limit,}}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = 0 \\

\]

Since, LHL=RHL, f(x) is continuous at x=0 and f(0)=0 .

For differentiability,

Left hand derivative

$

f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x - h} \right) - f\left( x \right)}}{{ - h}} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( { - h} \right) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \left( { - h} \right) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{ - h}} = - 1 \\

$

Right hand derivative

$

f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \\

f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( h \right) - 0}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{h} = 1 \\

$

Since, $LHD \ne RHD$ , f(x) is not differentiable at x=0

For applicability of Lagrange's Mean Value theorem on f(x) in interval $\left[ { - 1,1} \right]$ , Function must be continuous in $\left[ { - 1,1} \right]$ and differentiable in (1,1) .

But the function is not differentiable at x=0 .

So, Lagrange's Mean Value theorem is not applicable for f(x) in interval $\left[ { - 1,1} \right]$.

Note: Whenever we face such types of problems we use some important points. First we define the function on different intervals then check the continuity and differentiability of the function on different intervals. If function be continuous and differentiable on interval then Lagrange's mean value theorem be applicable.

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