Examine the application of Mean Value Theorem for all three functions given.
\[f\left( x \right)=\left[ x \right]forx\in \left[ -2,2 \right]\]

Hint: In order to check if we can apply the mean value theorem to the given function \[f\left( x \right)=\left[ x \right]forx\in \left[ -2,2 \right]\], firstly we must check for the continuity at every point and also at \[\left[ -2,2 \right]\]. Then differentiability at \[\left[ -2,2 \right]\] is to be checked. After checking the differentiability, we can conclude if we can apply the mean value theorem or not.

Complete step by step answer:
Now let us learn about the mean value theorem. It is one of the most important tools used in both differential and integral calculus. It helps us in identifying the behaviour at different functions. It is also called Lagrange's Mean Value Theorem. It’s statement states that \[f\left( x \right)\] is a function that satisfies the mentioned conditions. They are:
a.\[f\left( x \right)\] is continuous in \[\left[ a,b \right]\]
b. \[f\left( x \right)\] is differentiable in \[\left[ a,b \right]\].
There are two types of mean value theorem. They are: Cauchy’s mean value theorem and generalized mean value theorem.
Now let us start solving the given problem.
We are given with the function \[f\left( x \right)=\left[ x \right]forx\in \left[ -2,2 \right]\]
So we know that the given function \[f\left( x \right)\] is not continuous at every integral point.
Upon checking particularly, we can say that it is not continuous at \[x=2\] and \[x=-2\].
Now let us check for the differentiability.
It can be checked in the following way for \[f\] in \[\left[ -2,2 \right]\].
Let us consider an integer \[n\] such that \[n\in \left( -2,2 \right)\]
Now we will be checking the left hand limit of \[f\] at \[x=n\];
\[\Rightarrow \displaystyle \lim_{h \to {{0}^{-}}}\dfrac{f\left( n+h \right)-f\left( n \right)}{h}=\displaystyle \lim_{h \to {{0}^{-}}}\dfrac{\left[ n+h \right]-\left[ n \right]}{h}=\displaystyle \lim_{h \to {{0}^{-}}}\dfrac{n-1-n}{h}=\displaystyle \lim_{h \to {{0}^{-}}}\dfrac{-1}{h}=\infty \]
Now we will be checking the right hand limit of \[f\] at \[x=n\];
\[\Rightarrow \displaystyle \lim_{h \to {{0}^{+}}}\dfrac{f\left( n+h \right)-f\left( n \right)}{h}=\displaystyle \lim_{h \to {{0}^{+}}}\dfrac{\left[ n+h \right]-\left[ n \right]}{h}=\displaystyle \lim_{h \to {{0}^{+}}}\dfrac{n-n}{h}=\displaystyle \lim_{h \to {{0}^{+}}}0=0\]
We can clearly observe that the limit of \[f\] at \[x=n\] is not equal on the left hand and right hand.
So we can say that \[f\] is not differentiable at \[x=n\].
\[\therefore \] \[f\] is not differentiable in \[\left( -2,2 \right)\].
So we can observe that it does not satisfy the conditions of the mean value theorem. So the application of the mean value theorem is not applicable for the function \[f\left( x \right)=\left[ x \right]forx\in \left[ -2,2 \right]\].

Note: While solving for the mean value theorem we should definitely check for both the conditions. We must have a note of the conditions of the theorem. We can apply this theorem for increasing and decreasing nature of functions.