Answer is: option3

Solution:

The Mean Value Theorem states that if a function \( f(x) \) satisfies the following three conditions on an interval \([a, b]\):

1. Continuity on \([a, b]\): The function \( f(x) = -x^3 + 3x + 2 \) is continuous everywhere, including on \([0,3]\).
2. Differentiability on \( (a, b) \): The derivative of \( f(x) \), given by \( f'(x) = -3x^2 + 3 \), exists for all \( x \), so it is differentiable on \( (0,3) \).
3. Equal Function Values at Endpoints:
• \( f(0) = (-0)^3 + 3(0) + 2 = 2 \)
• \( f(3) = (-3)^3 + 3(3) + 2 = -27 + 9 + 2 = -16 \)

Since \( f(0) = f(3) \), all three conditions are satisfied.

Finding \( f'(c) \):

Using the Mean Value Theorem:

\[ f'(c) = \frac{f(3) - f(0)}{3 - 0} \]

Substituting values:

\[ f'(c) = \frac{-16 - 2}{3} = \frac{-18}{3} = -6 \]

Solving for \( c \):

We set \( f'(c) = -6 \):

\[ -3c^2 + 3 = -6 \]

Simplifying:

\[ -3c^2 = -9 \quad \Rightarrow \quad c^2 = 3 \]

\[ c = \pm \sqrt{3} \]

Checking the Valid \( c \) Values:

• \( \sqrt{3} \approx 1.732 \), which is in \( (0,3) \)
• \( -\sqrt{3} \approx -1.732 \), which is not in \( (0,3) \)

Final Answer: \( \sqrt{3} \)