Answer is: option3
\(\frac{3}{2}\)Solution:
Verify the conditions for MVT:
- \( V(x) = x^2 \) is continuous on \([0, 3]\).
- \( V(x) = x^2 \) is differentiable on \((0, 3)\).
Compute the average rate of change (slope of the secant line):
\[ \frac{V(3) - V(0)}{3 - 0} = \frac{9 - 0}{3} = 3 \]
Find the derivative of \( V(x) \):
\[ V'(x) = 2x \]
Set the derivative equal to the average rate of change and solve for \( c \):
\[ V'(c) = 2c = 3 \quad \Rightarrow \quad c = \frac{3}{2} \]
Check if \( c \) lies within the interval \( (0, 3) \):
\[ c = \frac{3}{2} \text{ is in } (0, 3) \]
Final Answer: \( \boxed{C} \)