Question

Check the applicability of Lagrange’s mean value theorem for $f(x) = \sqrt {5 - x} $ on $\left[ { - 3,6} \right]$.

Answer 1

Hint: Lagrange’s mean value theorem states that if there exists a function$f(x)$, such that:
1) $f$ is continuous on $\left[ {a,b} \right]$
2) $f$ is differentiable on $(a,b)$
 Then, there is at least one point $x = c,c \in (a,b)$ on this interval such that $f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$.

Complete step by step solution:
Firstly we should check if the function $f(x) = \sqrt {5 - x} $ is continuous or not.
We know that the values inside square root should be either equal to or greater than zero. Hence,
$
 5 - x \geqslant 0 \\
 5 \geqslant x \\
 $
Therefore, $f(x)$is defined only for $x \in ( - \infty ,5]$.
In the above question, $x \in [ - 3,6]$ and we just derived that the function is defined only till $5$ that implies that the function is not defined between $[5,6]$.
For example:
When $x = 6$
$
 f(x) = \sqrt {5 - 6} \\
 f(x) = \sqrt { - 1} \\
 $
Now, this value is undefined. Hence, the function is not defined for x=6.
Therefore, the function is not continuous on the domain $[ - 3,6]$.
Since, the first condition of Lagrange’s mean value theorem is not satisfied , hence we do not proceed with the other two criteria and we conclude that the function is not applicable for Lagrange's mean value theorem.

Note:
This theorem is also known as the first mean value theorem. On an interval it allows the increment of a function through the value of derivative at an intermediate point of the segment.