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?

Transcript text: 함수 $f(x)=\frac{1}{4} x^{3}-\frac{3}{2} x^{2}+1$ 에 대하여 $y=f(x)$ 의 그래프가 다음 그림과 같을 때, 함수 $f(x)$ 의 극댓값과 극솟값은?

Solution

Step 1: Find the first derivative of $ f(x) $

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) $.

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

Step 2: Set the first derivative equal to zero to find critical points

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:

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

Step 3: Determine the nature of the critical points using the second derivative

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.

The function $ f(x) $ has a local maximum at $ x = 0 $ and a local minimum at $ x = 4 $.

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