To find the local extrema of the function $f(x) = 3x + 2x^{-1}$, we start by finding the derivative and setting it to zero.

1. Find the derivative: $f'(x) = 3 - 2x^{-2} = 3 - \frac{2}{x^2}$

2. Set the derivative equal to zero: $3 - \frac{2}{x^2} = 0$ $\frac{2}{x^2} = 3$ $x^2 = \frac{2}{3}$ $x = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3}$

3. Determine whether these points are maxima or minima: We can use the second derivative test. First, compute the second derivative: $f''(x) = \frac{4}{x^3}$

   • For $x = \frac{\sqrt{6}}{3}$: $f''\left(\frac{\sqrt{6}}{3}\right) = \frac{4}{\left(\frac{\sqrt{6}}{3}\right)^3} > 0 \quad \text{(local minimum)}$

   • For $x = -\frac{\sqrt{6}}{3}$: $f''\left(-\frac{\sqrt{6}}{3}\right) = \frac{4}{\left(-\frac{\sqrt{6}}{3}\right)^3} < 0 \quad \text{(local maximum)}$

4. Find the function values at these critical points:

   • At $x = \frac{\sqrt{6}}{3}$: $f\left(\frac{\sqrt{6}}{3}\right) = 3\left(\frac{\sqrt{6}}{3}\right) + 2\left(\frac{3}{\sqrt{6}}\right) = \sqrt{6} + \frac{6}{\sqrt{6}} = \sqrt{6} + \sqrt{6} = 2\sqrt{6}$

   • At $x = -\frac{\sqrt{6}}{3}$: $f\left(-\frac{\sqrt{6}}{3}\right) = 3\left(-\frac{\sqrt{6}}{3}\right) + 2\left(-\frac{3}{\sqrt{6}}\right) = -\sqrt{6} - \sqrt{6} = -2\sqrt{6}$

Summary:

• Local maximum at $x = -\frac{\sqrt{6}}{3}$ with value $-2\sqrt{6}$.
• Local minimum at $x = \frac{\sqrt{6}}{3}$ with value $2\sqrt{6}$.

Do you want more details or have any questions? Here are some follow-up questions:

1. How do you interpret the second derivative test?
2. Can you explain the significance of local maxima and minima in real-world applications?
3. What would happen to the function if we changed the coefficients?
4. How do you graph this function to visualize the extrema?
5. What other methods can be used to find local extrema besides derivatives?

Tip: Always check for the domain of the function when finding extrema, especially with rational functions.