Answer
Verified
480.3k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE