Question $f(x)=x^{3}-3 x^{2}$ 的根大值点为（） P. 0. B. 2 C. -2 , D. 不有夻 $f(x)=x^{3}-3 x^{2}$ 的最大货点为

Studdy Solution

STEP 1

1. We are given the function $f(x) = x^3 - 3x^2$ .
2. We need to find the critical points of $f(x)$ to determine the maximum value.
3. Critical points occur where the derivative is zero or undefined.

STEP 2

1. Find the derivative of $f(x)$ .
2. Set the derivative equal to zero to find critical points.
3. Determine the nature of each critical point using the second derivative test.
4. Identify the maximum value and corresponding point.

STEP 3

Find the derivative of $f(x)$ :
$f'(x) = \frac{d}{dx}(x^3 - 3x^2)$

STEP 4

Calculate the derivative:
$f'(x) = 3x^2 - 6x$

STEP 5

Set the derivative equal to zero to find critical points:
$3x^2 - 6x = 0$

STEP 6

Factor the equation:
$3x(x - 2) = 0$

STEP 7

Solve for $x$ :
$x = 0 \quad \text{or} \quad x = 2$

STEP 8

Find the second derivative of $f(x)$ :
$f''(x) = \frac{d}{dx}(3x^2 - 6x)$

STEP 9

Calculate the second derivative:
$f''(x) = 6x - 6$

STEP 10

Evaluate the second derivative at each critical point to determine concavity:
For $x = 0$ :
$f''(0) = 6(0) - 6 = -6$
Since $f''(0) < 0$ , $x = 0$ is a local maximum.
For $x = 2$ :
$f''(2) = 6(2) - 6 = 6$
Since $f''(2) > 0$ , $x = 2$ is a local minimum.

STEP 11

Identify the maximum value and corresponding point:
The local maximum occurs at $x = 0$ .
The maximum value of $f(x)$ is at $x = 0$ .
$\boxed{0}$

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