F(x)=(2x-4)/(x^2-4x+8),omg=[3,6],find locol and globol minimum point

To find the local and global minimum points of the function \( F(x) = \frac{2x - 4}{x^2 - 4x + 8} \) over the interval \( [3, 6] \), we follow these steps:

### 1. **Differentiate the function**:
   We first need the derivative of \( F(x) \) to find critical points, where local extrema may occur.

\[
F'(x) = \frac{(x^2 - 4x + 8)'(2x - 4) - (2x - 4)'(x^2 - 4x + 8)}{(x^2 - 4x + 8)^2}
\]
We'll simplify that next.

### 2. **Solve for critical points**:
   Set \( F'(x) = 0 \) and solve for \( x \). This gives us potential critical points where the slope is zero.

### 3. **Check endpoints**:
   Since the interval is \( [3, 6] \), we also evaluate the function at the endpoints \( x = 3 \) and \( x = 6 \).

### 4. **Evaluate the function**:
   Once the critical points and endpoint values are known, we compare the values of \( F(x) \) at those points to find the local and global minima.

Let me proceed with these steps to find the exact solution. I'll calculate the derivative and solve it for you.The derivative of the function is:

\[
F'(x) = \frac{2x(4 - x)}{(x^4 - 8x^3 + 32x^2 - 64x + 64)}
\]

Setting \( F'(x) = 0 \), the critical points are \( x = 0 \) and \( x = 4 \). However, since we are only interested in the interval \( [3, 6] \), the critical point \( x = 4 \) is relevant.

### Next steps:
1. Evaluate \( F(x) \) at the critical point \( x = 4 \) and the endpoints \( x = 3 \) and \( x = 6 \).
2. Compare these values to determine the local and global minima.

Let's proceed with these calculations.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can manually evaluate the function \( F(x) = \frac{2x - 4}{x^2 - 4x + 8} \) at the points \( x = 3 \), \( x = 4 \), and \( x = 6 \) to compare the values.

Would you like further guidance on how to do this, or have any questions? Here are some related questions that might help:

1. How do we compute function values at specific points?
2. What is the role of the derivative in finding extrema?
3. Can we use second derivatives to confirm the nature of critical points?
4. How do we determine global minima and maxima over an interval?
5. What happens if there are no critical points within the interval?

**Tip**: Always check the function at both the critical points and the endpoints when looking for global extrema on a closed interval.

F(x) = (2x - 4) / (x^2 - 4x + 8), omg = [3, 6], find local and global minimum point

Learn how to find the local and global minimum points of the function F(x) = (2x - 4) / (x^2 - 4x + 8) over the interval [3, 6]. This solution includes key concepts such as differentiation, critical points, and the application of the first derivative test. Suitable for advanced high school calculus students.

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