F(x)=x^3-6x^2+9x-4,omg=(-1,5),find locol and globol minimum point

To find the **local** and **global** minimum points of the function \( F(x) = x^3 - 6x^2 + 9x - 4 \) on the interval \( (-1, 5) \), we can follow these steps:

### 1. Find the first derivative of the function.
The first derivative helps identify critical points where the slope of the tangent line is zero (possible minima or maxima).

\[
F'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x - 4) = 3x^2 - 12x + 9
\]

### 2. Set the first derivative equal to zero and solve for \( x \).
We solve the equation:

\[
3x^2 - 12x + 9 = 0
\]

This simplifies to:

\[
x^2 - 4x + 3 = 0
\]

Factoring the quadratic:

\[
(x - 3)(x - 1) = 0
\]

Thus, the critical points are \( x = 1 \) and \( x = 3 \).

### 3. Determine whether these points are minima, maxima, or neither.
To classify the critical points, we examine the second derivative or use the first derivative test.

- **Second derivative test**:
  
  The second derivative of \( F(x) \) is:

  \[
  F''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12
  \]

  - At \( x = 1 \):
  
    \[
    F''(1) = 6(1) - 12 = -6
    \]

    Since \( F''(1) < 0 \), \( x = 1 \) is a **local maximum**.

  - At \( x = 3 \):
  
    \[
    F''(3) = 6(3) - 12 = 6
    \]

    Since \( F''(3) > 0 \), \( x = 3 \) is a **local minimum**.

### 4. Evaluate the function at the critical points and the boundary points of the interval \( (-1, 5) \).

- At \( x = 1 \):
  
  \[
  F(1) = 1^3 - 6(1)^2 + 9(1) - 4 = 1 - 6 + 9 - 4 = 0
  \]

- At \( x = 3 \):
  
  \[
  F(3) = 3^3 - 6(3)^2 + 9(3) - 4 = 27 - 54 + 27 - 4 = -4
  \]

- At the boundary point \( x = -1 \):
  
  \[
  F(-1) = (-1)^3 - 6(-1)^2 + 9(-1) - 4 = -1 - 6 - 9 - 4 = -20
  \]

- At the boundary point \( x = 5 \):
  
  \[
  F(5) = 5^3 - 6(5)^2 + 9(5) - 4 = 125 - 150 + 45 - 4 = 16
  \]

### 5. Compare the function values.
- \( F(-1) = -20 \)
- \( F(1) = 0 \)
- \( F(3) = -4 \)
- \( F(5) = 16 \)

### Conclusion:
- The **global minimum** on the interval \( (-1, 5) \) occurs at \( x = -1 \), where \( F(x) = -20 \).
- The **local minimum** occurs at \( x = 3 \), where \( F(x) = -4 \).

Let me know if you need more details or have any questions!

Here are some related questions:
1. How do you find local maxima and minima using the first derivative test?
2. What is the significance of the second derivative in determining concavity?
3. Can a function have more than one global minimum or maximum?
4. How do boundary points affect global extrema in a closed interval?
5. What are other methods to find global extrema besides evaluating critical and boundary points?

**Tip**: Always check the boundary points when looking for global extrema on a closed interval, as the minimum or maximum may occur there.

F(x) = x^3 - 6x^2 + 9x - 4, omg = (-1, 5), find local and global minimum point

Explore how to find the local and global minimum points of the function F(x) = x^3 - 6x^2 + 9x - 4 on the interval (-1, 5). This solution uses calculus concepts like derivatives, critical points, and the second derivative test, suitable for Grades 11-12.

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