Questions: 함수 f(x)=1/4 x^3 - 3/2 x^2 + 1 에 대하여 y=f(x) 의 그래프가 다음 그림과 같을 때, 함수 f(x) 의 극댓값과 극솟값은?

Solution Steps

Given the function \( f(x) = \frac{1}{4}x^3 - \frac{3}{2}x^2 + 1 \), we need to find the first derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx} \left( \frac{1}{4}x^3 - \frac{3}{2}x^2 + 1 \right) \] \[ f'(x) = \frac{3}{4}x^2 - 3x \]

To find the critical points, we set \( f'(x) = 0 \).

\[ \frac{3}{4}x^2 - 3x = 0 \] \[ \frac{3}{4}x(x - 4) = 0 \]

This gives us two solutions:

\[ x = 0 \] \[ x = 4 \]

To determine whether these critical points are maxima, minima, or points of inflection, we find the second derivative \( f''(x) \).

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

Evaluate \( f''(x) \) at the critical points:

For \( x = 0 \): \[ f''(0) = \frac{3}{2}(0) - 3 = -3 \] Since \( f''(0) < 0 \), \( x = 0 \) is a local maximum.

For \( x = 4 \): \[ f''(4) = \frac{3}{2}(4) - 3 = 6 - 3 = 3 \] Since \( f''(4) > 0 \), \( x = 4 \) is a local minimum.

Final Answer