F(x)=x^3-3x^2+2x find tge local maxima and minima

To find the local maxima and minima of the function \( F(x) = x^3 - 3x^2 + 2x \), we need to follow these steps:

### Step 1: Find the first derivative of \( F(x) \)
The first derivative will help us find the critical points where the slope is zero.

\[
F'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2x)
\]
\[
F'(x) = 3x^2 - 6x + 2
\]

### Step 2: Set the first derivative equal to zero to find critical points
\[
3x^2 - 6x + 2 = 0
\]

Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)}
\]
\[
x = \frac{6 \pm \sqrt{36 - 24}}{6}
\]
\[
x = \frac{6 \pm \sqrt{12}}{6}
\]
\[
x = \frac{6 \pm 2\sqrt{3}}{6}
\]
\[
x = 1 \pm \frac{\sqrt{3}}{3}
\]

Thus, the critical points are:
\[
x_1 = 1 + \frac{\sqrt{3}}{3}, \quad x_2 = 1 - \frac{\sqrt{3}}{3}
\]

### Step 3: Use the second derivative to classify the critical points
The second derivative will help us determine whether each critical point is a local maximum or minimum.

\[
F''(x) = \frac{d}{dx}(3x^2 - 6x + 2)
\]
\[
F''(x) = 6x - 6
\]

Now, evaluate \( F''(x) \) at each critical point:

1. For \( x_1 = 1 + \frac{\sqrt{3}}{3} \):
   \[
   F''(x_1) = 6\left(1 + \frac{\sqrt{3}}{3}\right) - 6 = 6 + 2\sqrt{3} - 6 = 2\sqrt{3}
   \]
   Since \( F''(x_1) > 0 \), \( x_1 \) is a **local minimum**.

2. For \( x_2 = 1 - \frac{\sqrt{3}}{3} \):
   \[
   F''(x_2) = 6\left(1 - \frac{\sqrt{3}}{3}\right) - 6 = 6 - 2\sqrt{3} - 6 = -2\sqrt{3}
   \]
   Since \( F''(x_2) < 0 \), \( x_2 \) is a **local maximum**.

### Conclusion:
- Local minimum at \( x_1 = 1 + \frac{\sqrt{3}}{3} \).
- Local maximum at \( x_2 = 1 - \frac{\sqrt{3}}{3} \).

Would you like more details or further clarification?

Here are 5 related questions to explore:

1. How do you find inflection points of a function?
2. What is the significance of the second derivative in classifying critical points?
3. Can local maxima and minima occur at endpoints of a function's domain?
4. What are the global maximum and minimum values of a function?
5. How does the behavior of cubic functions differ from quadratic functions in terms of critical points?

**Tip:** Always check the second derivative at critical points to confirm whether they are maxima or minima.

F(x) = x^3 - 3x^2 + 2x find the local maxima and minima

Explore how to find the local maxima and minima for the cubic function F(x) = x^3 - 3x^2 + 2x using the first and second derivative tests. This detailed solution walks through critical points and classifying them as local maxima or minima, ideal for calculus students in Grades 11-12.

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