Examine if Rolle's Theorem is applicable to the function $f(x) = x^{2} - 1$ for $x \in [1, 2]$. Can you say something about the converse of Rolle's Theorem from this example?

(N/A) According to Rolle's Theorem,for a function $f: [a, b] \rightarrow \mathbb{R}$,if:
$1$) $f$ is continuous on $[a, b]$
$2$) $f$ is differentiable on $(a, b)$
$3$) $f(a) = f(b)$
Then,there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
For the function $f(x) = x^{2} - 1$ on the interval $[1, 2]$:
- $f(x)$ is a polynomial function,so it is continuous on $[1, 2]$ and differentiable on $(1, 2)$.
- Calculate the values at the endpoints:
$f(1) = (1)^{2} - 1 = 0$
$f(2) = (2)^{2} - 1 = 3$
- Since $f(1) \neq f(2)$,the third condition of Rolle's Theorem is not satisfied.
Therefore,Rolle's Theorem is not applicable to $f(x) = x^{2} - 1$ on $[1, 2]$.
Regarding the converse: The converse of Rolle's Theorem states that if there exists $c \in (a, b)$ such that $f'(c) = 0$,then $f(a) = f(b)$. This is not necessarily true. For example,if $f(x) = x^{2}$,$f'(0) = 0$ at $x = 0$,but $f(-1) = 1$ and $f(1) = 1$. However,for other functions,$f'(c) = 0$ does not imply $f(a) = f(b)$.