Let's solve this step by step.

1. Rolle's Theorem Conditions

Rolle's Theorem can be applied if the following conditions are met:

• The function $f(x)$ is continuous on the closed interval $[a, b]$.
• The function $f(x)$ is differentiable on the open interval $(a, b)$.
• $f(a) = f(b)$.

Here, the function is $f(x) = -x^2 + 9x$ on the interval $[0, 9]$.

Step 1: Check if $f(x)$ is continuous on $[0, 9]$

$f(x) = -x^2 + 9x$ is a polynomial function, and polynomial functions are continuous everywhere. Therefore, $f(x)$ is continuous on the closed interval $[0, 9]$.

Step 2: Check if $f(x)$ is differentiable on $(0, 9)$

Polynomial functions are also differentiable everywhere. Hence, $f(x)$ is differentiable on the open interval $(0, 9)$.

Step 3: Check if $f(0) = f(9)$

• $f(0) = -(0)^2 + 9(0) = 0$
• $f(9) = -(9)^2 + 9(9) = -81 + 81 = 0$

Since $f(0) = f(9) = 0$, the third condition is also satisfied.

Therefore, Rolle's Theorem can be applied.

2. Finding $c$ such that $f'(c) = 0$

Next, we need to find the value(s) of $c$ in the open interval $(0, 9)$ where $f'(c) = 0$.

• First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(-x^2 + 9x) = -2x + 9$

• Now, set $f'(x) = 0$ and solve for $x$: $-2x + 9 = 0$ $2x = 9$ $x = \frac{9}{2} = 4.5$

Thus, $c = 4.5$ is the value where $f'(c) = 0$.

Answer

• Rolle's Theorem can be applied.
• The value of $c$ is $4.5$.

Would you like more details or have any questions?

Here are some related questions:

1. What are the key conditions for applying Rolle's Theorem?
2. How do you prove that a polynomial is continuous and differentiable?
3. Can you apply Rolle’s Theorem if the function is not differentiable at a point in the interval?
4. What is the geometric interpretation of the value $c$ in Rolle’s Theorem?
5. How would you approach a function with different behavior (e.g., piecewise) on the interval?

Tip: Rolle's Theorem is a special case of the Mean Value Theorem where $f(a) = f(b)$, ensuring a horizontal tangent line at some point in the interval.