To find the local minimum and maximum of the function $f(x) = 5x + 3x^{-1}$, we can follow these steps:

Step 1: Compute the Derivative of $f(x)$

The first step is to find $f'(x)$ and set it to zero to locate the critical points.

Given: $f(x) = 5x + \frac{3}{x}$

The derivative $f'(x)$ is: $f'(x) = 5 - \frac{3}{x^2}$

Step 2: Set the Derivative Equal to Zero

To find the critical points: $5 - \frac{3}{x^2} = 0$

Rearrange to solve for $x$: $5 = \frac{3}{x^2} \Rightarrow x^2 = \frac{3}{5} \Rightarrow x = \pm \sqrt{\frac{3}{5}}$

Thus, we have two critical points: $x = \sqrt{\frac{3}{5}} \quad \text{and} \quad x = -\sqrt{\frac{3}{5}}$

Step 3: Determine if Each Critical Point is a Maximum or Minimum

To classify these points, we can use the second derivative test.

Second Derivative $f''(x)$

The second derivative of $f(x)$ is: $f''(x) = \frac{6}{x^3}$

Now, we evaluate $f''(x)$ at each critical point:

1. At $x = \sqrt{\frac{3}{5}}$: $f''\left(\sqrt{\frac{3}{5}}\right) = \frac{6}{\left(\sqrt{\frac{3}{5}}\right)^3} > 0$ Since $f''(x) > 0$, this point is a local minimum.

2. At $x = -\sqrt{\frac{3}{5}}$: $f''\left(-\sqrt{\frac{3}{5}}\right) = \frac{6}{\left(-\sqrt{\frac{3}{5}}\right)^3} < 0$ Since $f''(x) < 0$, this point is a local maximum.

Step 4: Compute the Function Values at the Critical Points

To find the values of $f(x)$ at these points, substitute back into the original function.

1. Local Maximum at $x = -\sqrt{\frac{3}{5}}$: $f\left(-\sqrt{\frac{3}{5}}\right) = 5\left(-\sqrt{\frac{3}{5}}\right) + \frac{3}{-\sqrt{\frac{3}{5}}}$

2. Local Minimum at $x = \sqrt{\frac{3}{5}}$: $f\left(\sqrt{\frac{3}{5}}\right) = 5\left(\sqrt{\frac{3}{5}}\right) + \frac{3}{\sqrt{\frac{3}{5}}}$

After solving these expressions, you'll find the exact values for the local maximum and minimum. Let me know if you'd like further calculations!

---

Do you need the explicit values calculated or have any questions on the process? Here are some related questions for practice:

1. What is the process for finding inflection points of a function?
2. How does the second derivative test determine concavity?
3. What are the potential applications of finding local extrema in real-world problems?
4. How can critical points be used in optimization problems?
5. What does it mean if the second derivative is zero at a critical point?

Tip: Always verify if a critical point is within the domain of the function, as this affects its validity for extrema classification.