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

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

### 1. Find the first derivative
We start by finding the first derivative \( F'(x) \) to locate the critical points.

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

### 2. Solve \( F'(x) = 0 \) to find critical points
Now, we solve the equation \( 3x^2 - 12x + 9 = 0 \) to find the critical points.

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

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

Factoring the quadratic equation:

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

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

### 3. Evaluate the second derivative for concavity
The second derivative test will help determine if these critical points correspond to a local minimum, maximum, or inflection point.

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

- For \( x = 1 \), \( F''(1) = 6(1) - 12 = -6 \), which is negative, indicating that \( x = 1 \) is a local maximum.
- For \( x = 3 \), \( F''(3) = 6(3) - 12 = 6 \), which is positive, indicating that \( x = 3 \) is a local minimum.

### 4. Evaluate the function at critical points and endpoints
To find the global minimum, we evaluate the function at the critical points \( x = 1 \), \( x = 3 \), and the endpoints of the interval \( x = -1 \) and \( x = 5 \).

\[
F(1) = (1)^3 - 6(1)^2 + 9(1) - 4 = 1 - 6 + 9 - 4 = 0
\]
\[
F(3) = (3)^3 - 6(3)^2 + 9(3) - 4 = 27 - 54 + 27 - 4 = -4
\]
\[
F(-1) = (-1)^3 - 6(-1)^2 + 9(-1) - 4 = -1 - 6 - 9 - 4 = -20
\]
\[
F(5) = (5)^3 - 6(5)^2 + 9(5) - 4 = 125 - 150 + 45 - 4 = 16
\]

### 5. Conclusion: Local and global minima
- The local minimum occurs at \( x = 3 \), with \( F(3) = -4 \).
- The global minimum over the interval \( (-1, 5) \) is at \( x = -1 \), with \( F(-1) = -20 \).

Would you like more details or explanations on any step?

### Further exploration:
1. What does it mean for a function to have a local minimum versus a global minimum?
2. How do you interpret the second derivative test?
3. Why do we evaluate endpoints for global extrema?
4. What happens if the interval is unbounded?
5. What are some common functions with multiple local extrema?

### Tip:
When finding global extrema, always check the function's value at the endpoints of the given interval!

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

Discover how to find the local and global minimum of the cubic function F(x) = x^3 - 6x^2 + 9x - 4 on the interval (-1,5). This step-by-step guide covers critical points, the first and second derivative tests, and endpoint evaluations. Ideal for Grades 11-12 or introductory calculus students.

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