Answer is: option1

Solution:

Find the first derivative \( f'(x) \):

\[ f(x) = \frac{x^4}{2} - \frac{x^5}{10} \]

\[ f'(x) = \frac{d}{dx} \left( \frac{x^4}{2} \right) - \frac{d}{dx} \left( \frac{x^5}{10} \right) = 2x^3 - \frac{x^4}{2} \]

Find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx} (2x^3) - \frac{d}{dx} \left( \frac{x^4}{2} \right) = 6x^2 - 2x^3 \]

Determine critical points of \( f'(x) \) by setting \( f''(x) = 0 \):

\[ 6x^2 - 2x^3 = 0 \]

\[ 2x^2(3 - x) = 0 \]

\[ x = 0 \quad \text{or} \quad x = 3 \]

Analyze the critical points:

• For \( x < 0 \) (e.g., \( x = -1 \)):
  \( f''(-1) = 6(-1)^2 - 2(-1)^3 = 6 + 2 = 8 > 0 \)
• For \( 0 < x < 3 \) (e.g., \( x = 1 \)):
  \( f''(1) = 6(1)^2 - 2(1)^3 = 6 - 2 = 4 > 0 \)
• For \( x > 3 \) (e.g., \( x = 4 \)):
  \( f''(4) = 6(4)^2 - 2(4)^3 = 96 - 128 = -32 < 0 \)

The second derivative changes from positive to negative at \( x = 3 \), indicating that \( f'(x) \) has a maximum at \( x = 3 \).

Conclusion: The derivative \( f'(x) \) attains its maximum value at \( x = 3 \).

Answer: A