Determine whether Rolle's Theorem can be applied to the function $f ( x ) = x ^ { 2 } - 2 x - 3$ on the closed interval $( - 1,3 )$. If Rolle's Theorem can be applied, find all values of $c$ in the open interval $( - 1,3 )$ such that $f ^ { t } ( c ) = 0$.
A) Rolle's Theorem applies; $c = 1$
B) Rolle's Theorem applies; $\mathrm { c } = 2$
C) Rolle's Theorem applies; $c = 0$
D) Rolle's Theorem applies; $c = - 1$
E) Rolle's Theorem does not apply

Question

Quizplus · Accepted Answer

For Rolle's Theorem to apply, the function must be continuous on the closed interval $( - 1,3 )$ and differentiable on the open interval $ ( - 1,3 )$. 
$f(x) = x^2 - 2x - 3$ is a polynomial function and is therefore continuous and differentiable for all real numbers.
Furthermore, $f(-1) = 6$ and $f(3)=-6$ so the function takes on different values at the endpoints of the interval. By Rolle's Theorem, there exists at least one point $c$ in the open interval $( - 1,3 )$ such that $f'(c) = 0$.
Now, 
$$f(x) = x^2 - 2x - 3$$
$$\Rightarrow f'(x) = 2x -2$$
Setting $f'(x) = 0$ and solving for $x$ gives us 
$$2x-2 = 0 \Rightarrow x = 1$$
Since $x = 1$ lies in the interval $( - 1,3 )$, we can conclude that Rolle's Theorem applies and $c = 1$ is the only point in the open interval $( - 1,3 )$ such that $f'(c) = 0$. Thus, the answer is A.

Determine Whether Rolle's Theorem Can Be Applied to the Function f(x)=x2−2x−3f ( x ) = x ^ { 2 } - 2 x - 3f(x)=x2−2x−3

Determine Whether Rolle's Theorem Can Be Applied to the Function $f ( x ) = x ^ { 2 } - 2 x - 3$