The function f(x)=8x+2x^-1 has one local minimum and one local maximum.
This function has a local maximum at x=
with value

and a local minimum at x=
with value

To find the local maximum and minimum of the function \( f(x) = 8x + 2x^{-1} \), let's go through the steps:

1. **Find the derivative** \( f'(x) \) to locate critical points:
   \[
   f(x) = 8x + 2x^{-1}
   \]
   Differentiating with respect to \( x \):
   \[
   f'(x) = 8 - 2x^{-2} = 8 - \frac{2}{x^2}
   \]

2. **Set \( f'(x) = 0 \)** to find critical points:
   \[
   8 - \frac{2}{x^2} = 0
   \]
   Solving for \( x \):
   \[
   \frac{2}{x^2} = 8
   \]
   \[
   x^2 = \frac{2}{8} = \frac{1}{4}
   \]
   \[
   x = \pm \frac{1}{2}
   \]

   So, the critical points are \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

3. **Determine if these points are maxima or minima** using the second derivative \( f''(x) \):
   \[
   f''(x) = \frac{4}{x^3}
   \]
   
   - At \( x = \frac{1}{2} \):
     \[
     f''\left(\frac{1}{2}\right) = \frac{4}{\left(\frac{1}{2}\right)^3} = \frac{4}{\frac{1}{8}} = 32 > 0
     \]
     Since \( f''(x) > 0 \), \( x = \frac{1}{2} \) is a local minimum.

   - At \( x = -\frac{1}{2} \):
     \[
     f''\left(-\frac{1}{2}\right) = \frac{4}{\left(-\frac{1}{2}\right)^3} = \frac{4}{-\frac{1}{8}} = -32 < 0
     \]
     Since \( f''(x) < 0 \), \( x = -\frac{1}{2} \) is a local maximum.

4. **Calculate the function values at these points**:

   - For the local minimum at \( x = \frac{1}{2} \):
     \[
     f\left(\frac{1}{2}\right) = 8 \cdot \frac{1}{2} + 2 \cdot \frac{1}{\frac{1}{2}} = 4 + 4 = 8
     \]

   - For the local maximum at \( x = -\frac{1}{2} \):
     \[
     f\left(-\frac{1}{2}\right) = 8 \cdot \left(-\frac{1}{2}\right) + 2 \cdot \frac{1}{-\frac{1}{2}} = -4 - 4 = -8
     \]

### Summary of Results:
- The local maximum occurs at \( x = -\frac{1}{2} \) with value \( -8 \).
- The local minimum occurs at \( x = \frac{1}{2} \) with value \( 8 \).

Would you like a more detailed explanation on any of these steps?

---

1. What is the significance of the second derivative test for identifying maxima and minima?
2. How would the result change if we modified the function to \( f(x) = 8x + \frac{3}{x} \)?
3. What happens to \( f(x) \) as \( x \) approaches 0?
4. How do we interpret the local maxima and minima on a graph?
5. Can this method be applied to find inflection points?

**Tip:** When differentiating functions with terms like \( x^{-1} \), rewriting them as fractions or powers simplifies the derivative process.

The function f(x)=8x+2x^-1 has one local minimum and one local maximum. This function has a local maximum at x= with value and a local minimum at x= with value

Explore how to find the local maximum and minimum of the function f(x) = 8x + 2x^-1 using differentiation and the second derivative test. This problem includes step-by-step calculations of derivatives and critical points, ideal for calculus students in Grades 11-12.

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