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

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

### Step 1: **Find the critical points (local extrema)**

1. **Find the derivative \( F'(x) \):**
   We apply the quotient rule to differentiate the function.
   \[
   F'(x) = \frac{(x^2 - 4x + 8)'(2x - 4) - (x^2 - 4x + 8)(2x - 4)'}{(x^2 - 4x + 8)^2}
   \]
   Let me calculate this.

### Step 2: **Check the boundaries for global minimum**

Evaluate \( F(x) \) at the boundary points \( x = 3 \) and \( x = 6 \).

### Step 3: **Compare values at critical points and boundary points to find the global minimum**

Let me calculate everything for you.### Step 1: **Critical Points and Values**
- The critical points are \( x = 0 \) and \( x = 4 \).
- The function values at these critical points are:
  - \( F(0) = -\frac{1}{2} \)
  - \( F(4) = \frac{1}{2} \)

### Step 2: **Boundary Points and Values**
- The boundary points are \( x = 3 \) and \( x = 6 \).
- The function values at these boundary points are:
  - \( F(3) = \frac{2}{5} \)
  - \( F(6) = \frac{2}{5} \)

### Step 3: **Determine Local and Global Minimum**
- The smallest function value occurs at \( x = 0 \), where \( F(0) = -\frac{1}{2} \). This represents the **global minimum** since it's the lowest value.
- The other points (\( x = 3 \), \( x = 4 \), and \( x = 6 \)) yield higher values, with \( F(3) = F(6) = \frac{2}{5} \) and \( F(4) = \frac{1}{2} \).

However, since \( x = 0 \) is not in the interval \([3, 6]\), we consider the values within the interval. So, within \([3, 6]\):
- The minimum value is \( F(3) = F(6) = \frac{2}{5} \), at both \( x = 3 \) and \( x = 6 \).

### Conclusion:
- **Local minimum** within \([3, 6]\): \( F(3) = F(6) = \frac{2}{5} \) at \( x = 3 \) and \( x = 6 \).
- **Global minimum** on \([3, 6]\): \( \frac{2}{5} \) at both boundaries.

Would you like more details, or have any questions?

Here are 5 follow-up questions to expand on this:

1. How do you apply the second derivative test to check the nature of critical points?
2. What is the significance of boundary points in finding global minima/maxima?
3. How does the quotient rule work in calculating derivatives of rational functions?
4. Can a function have multiple global minima on a closed interval?
5. How would the behavior of the function change if the interval were expanded beyond \([3, 6]\)?

**Tip**: Always check both the critical points and boundaries when dealing with closed intervals!

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

Learn how to find the local and global minimum of the function F(x) = (2x - 4) / (x^2 - 4x + 8) on the interval [3, 6]. This step-by-step solution involves calculus concepts like critical points, the quotient rule for differentiation, and boundary evaluations.

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