Question

Discuss the applicability of Lagrange's mean value theorem for the function $f\left( x \right) = \mid x\mid \;$ on [−1,1 ] .

Answer 1

In order to solve this question we need to know the concept of Lagrange's mean value theorem The Mean Value Theorem states that for any given curve between two endpoints, there must be a point at which the slope of the tangent to the curve is same as the slope of the secant through its endpoints.
If f(x) is a function, so that f(x) is continuous on the closed interval [a, b] and also differentiable on the open interval (a, b), then there is point c in (a, b) that is, a < c < b such that
$f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$

Complete step by step solution:
As per the new definition, we can claim just by seeing that the function is continuous and differentiable everywhere except 0 (to be checked ).

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

Firstly the Left Hand Derivative of f(x) at
$x = 0 \to {\lim _{h \to 0}}\dfrac{{f(0 - h) - f(0)}}{{(0 - h) - 0}} \\\ = {\lim _{h \to 0}}\dfrac{{f(0 - h) - f(0)}}{{(0 - h) - 0}} = - 1 \\\$
Secondly the Right Hand Derivative of f(x) at
$x = 0 \to {\lim _{h \to 0}}\dfrac{{f(0 + h) - f(0)}}{{(0 + h) - 0}} \\\ = {\lim _{h \to 0}}\dfrac{{f(0 + h) - f(0)}}{{(0 + h) - 0}} = 1 \\\$
Since , the Left Hand Derivative = -1 and the Right Hand Derivative = 1 that is Left Hand Derivative is not equal to Right Hand Derivative , f(x) is not differentiable at x=0 .
But to be valid or applicable for the Lagrange's mean value theorem on f(x) in interval [−1,1 ] , The function must be continuous in [ -1,1 ] and differentiable in ( -1,1 ) .
Therefore , Lagrange's mean value theorem is not applicable for f ( x ) in interval [ -1,1 ]

Note: There can be more than one such c.
We can also simply call it the first mean value theorem so , do not confuse the two terms .
The Lagrange's mean value theorem is applicable only if the f ( x ) is continuous on [ -1,1 ] and differentiable on ( -1 , 1 ) for the above given question .