Examine the applicability of the Mean Value Theorem for the following functions:
$(i)$ $f(x) = [x]$ for $x \in [5, 9]$
$(ii)$ $f(x) = [x]$ for $x \in [-2, 2]$
$(iii)$ $f(x) = x^{2} - 1$ for $x \in [1, 2]$

(N/A) The Mean Value Theorem states that for a function $f: [a, b] \rightarrow \mathbb{R}$,if:
a) $f$ is continuous on $[a, b]$
b) $f$ is differentiable on $(a, b)$
Then,there exists some $c \in (a, b)$ such that $f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}$.
$(i)$ $f(x) = [x]$ for $x \in [5, 9]$. The greatest integer function $[x]$ is not continuous at any integer point. Since $[5, 9]$ contains integers,$f(x)$ is not continuous on $[5, 9]$. Thus,the Mean Value Theorem is not applicable.
$(ii)$ $f(x) = [x]$ for $x \in [-2, 2]$. Similar to $(i)$,the function is not continuous at integer points like $-1, 0, 1$. Thus,the Mean Value Theorem is not applicable.
$(iii)$ $f(x) = x^{2} - 1$ for $x \in [1, 2]$. $f(x)$ is a polynomial function,which is continuous on $[1, 2]$ and differentiable on $(1, 2)$. Thus,the Mean Value Theorem is applicable.
We find $c$ such that $f^{\prime}(c) = \frac{f(2) - f(1)}{2 - 1} = \frac{3 - 0}{1} = 3$. Since $f^{\prime}(x) = 2x$,we have $2c = 3$,which gives $c = 1.5$. Since $1.5 \in (1, 2)$,the theorem holds.