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.