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

To find the minimum of the function \( F(x) = \frac{2x - 4}{x^2 - 4x + 8} \) on the interval \([3,6]\), we need to analyze both the critical points and the behavior of the function at the endpoints.

### Steps:
1. **Find the first derivative** \( F'(x) \) to identify the critical points.
2. **Evaluate the critical points** within the interval \([3, 6]\).
3. **Check the function's values** at the endpoints \( x = 3 \) and \( x = 6 \) for the global minimum.
4. Compare these values to determine the minimum local and global points.

#### Step 1: Find the first derivative of \( F(x) \)
The derivative \( F'(x) \) of the given function can be found using the quotient rule:
\[
F'(x) = \frac{(g(x)h'(x) - g'(x)h(x))}{h(x)^2}
\]
where \( g(x) = 2x - 4 \) and \( h(x) = x^2 - 4x + 8 \).

Let me compute the derivative.The first derivative of the function \( F(x) = \frac{2x - 4}{x^2 - 4x + 8} \) is:

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

#### Step 2: Critical Points
To find critical points, set \( F'(x) = 0 \) and solve for \( x \).

The numerator \( 2x(4 - x) \) is equal to zero when:
\[
2x(4 - x) = 0
\]
This gives the solutions \( x = 0 \) and \( x = 4 \).

Now, let's check which of these points lie within the interval \([3, 6]\). Since \( x = 4 \) is in the interval, it is a critical point of interest.

#### Step 3: Evaluate the function at critical points and endpoints
We need to evaluate \( F(x) \) at \( x = 3 \), \( x = 4 \), and \( x = 6 \) to determine the minimum.

Let me compute the values of \( F(x) \) at these points.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like me to guide you through the manual calculation of \( F(x) \) at these points or clarify anything else? Let me know if you'd like more assistance.

Meanwhile, here are some related questions for further understanding:
1. How do we classify critical points as local minima or maxima?
2. What is the importance of endpoints when determining global extrema?
3. Can we always find critical points using the first derivative?
4. How does the second derivative test help in identifying the nature of critical points?
5. What are the possible issues when solving rational functions?

**Tip**: Always check both critical points and endpoints when looking for a global minimum or maximum on a closed interval.

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

Discover how to find the local and global minimum of the rational function F(x) = (2x-4)/(x^2-4x+8) on the interval [3,6]. This calculus problem involves critical points, the first derivative, and global extrema evaluation. Suitable for advanced high school and college students.

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