Determine whether Rolle's theorem can be applied to $$ f $$ on the closed interval $$ [a, b] $$. (Select all that apply.) $$ f(x)=-x^{2}+6 x, \quad[0,6] $$ $$ \square $$ Yes, Rolle's theorem can be applied. $$ \square $$ No, because $$ f $$ is not continuous on the closed interval $$ [a, b] $$. $$ \square $$ No, because $$ f $$ is not differentiable on the open interval $$ (a, b) $$. $$ \square $$ No, because $$ f(a) \neq \mathbb{P} b) $$. If Rolle's theorem can be applied, find all values of $$ \epsilon $$ in the open interval $$ (a, b) $$ such that $$ f^{\prime}(c)=0 $$. (Enter your answers as a comma-separated list. If Rolle's theorem cannot be applied, enter NA.) $$ c=\square $$

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We first check the conditions of Rolle's theorem for $$ f(x)=-x^2+6x $$ on the interval $$[0,6]$$:

1. Since $$ f $$ is a polynomial, it is continuous everywhere, and in particular on $$[0,6]$$.
2. Again, as a polynomial, $$ f $$ is differentiable everywhere, and hence differentiable on $$ (0,6) $$.
3. We need $$ f(0)=f(6) $$. Compute:
   $$
   f(0) = -(0)^2+6(0)=0,
   $$
   $$
   f(6) = -(6)^2+6(6)=-36+36=0.
   $$
   So, $$ f(0)=f(6) $$.

Since all three conditions are met, Rolle's theorem can be applied.

Next, by Rolle's theorem, there exists at least one $$ c $$ in $$ (0,6) $$ such that $$ f'(c)=0 $$. We find $$ c $$ as follows:

Compute the derivative:
$$
f'(x) = \frac{d}{dx}(-x^2+6x)=-2x+6.
$$
Set the derivative equal to zero:
$$
-2x+6=0.
$$
Solve for $$ x $$:
$$
-2x=-6 \quad\Longrightarrow\quad x=3.
$$

Thus, the only value of $$ c $$ in $$ (0,6) $$ such that $$ f'(c)=0 $$ is:
$$
c=3.
$$

The correct selections are:

- $$ \square $$ Yes, Rolle's theorem can be applied.
- $$ \square $$ No, because $$ f $$ is not continuous on the closed interval $$ [a, b] $$.  (Incorrect)
- $$ \square $$ No, because $$ f $$ is not differentiable on the open interval $$ (a, b) $$.  (Incorrect)
- $$ \square $$ No, because $$ f(a) \neq f(b) $$.  (Incorrect)

And the answer for $$ c $$ is:
$$
c=3.
$$
Rolle's theorem can be applied. There is one value of $$ c $$ in $$ (0,6) $$ where $$ f'(c)=0 $$, which is $$ c=3 $$.

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