Question

Знайти критичні точки функції f(x)=3x в кубе-x квадрат пожалуйста помогитее!!срочно очень нужно!!!

Answer 1

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Answer 2

Finding Critical Points of the Function f(x) = 3x^3 - x^2

To find the critical points of a function, we need to find the values of x where the derivative of the function is equal to zero or undefined. The critical points can be local minimums, local maximums, or inflection points.

Let's find the critical points of the function f(x) = 3x^3 - x^2 step by step:

1. Find the derivative of the function f(x) with respect to x: ``` f'(x) = 9x^2 - 2x ```

2. Set the derivative equal to zero and solve for x: ``` 9x^2 - 2x = 0 ```

Factoring out x, we get: ``` x(9x - 2) = 0 ```

Setting each factor equal to zero, we have two possibilities: ``` x = 0 or 9x - 2 = 0 ```

Solving the second equation, we find: ``` 9x = 2 x = 2/9 ```

So the critical points are x = 0 and x = 2/9.

3. To determine the nature of these critical points (whether they are local minimums, local maximums, or inflection points), we need to analyze the second derivative.

Taking the derivative of f'(x), we get: ``` f''(x) = 18x - 2 ```

4. Substitute the critical points into the second derivative to determine their nature:

For x = 0: ``` f''(0) = 18(0) - 2 = -2 ```

For x = 2/9: ``` f''(2/9) = 18(2/9) - 2 = 4 - 2 = 2 ```

Based on the second derivative test, we can conclude the following:

- When f''(x) < 0, the function has a local maximum at that point. - When f''(x) > 0, the function has a local minimum at that point. - When f''(x) = 0, the test is inconclusive.

Therefore, at x = 0, the function has a local maximum, and at x = 2/9, the function has a local minimum.

Summary:

The critical points of the function f(x) = 3x^3 - x^2 are x = 0 (local maximum) and x = 2/9 (local minimum) [[1]].