To analyze the function $f(x) = 15x + \frac{5}{x}$ and answer the questions, we need to find the derivative and determine where it is positive, negative, or zero. This will help us identify where the function is increasing, decreasing, and where any local maxima or minima are located.

Step 1: Find $f'(x)$

The first derivative $f'(x)$ tells us about the slope of $f(x)$. We calculate $f'(x)$ as follows: $f(x) = 15x + \frac{5}{x}$ Using the power rule and the derivative of $\frac{1}{x}$, we get: $f'(x) = 15 - \frac{5}{x^2}$

Step 2: Set $f'(x) = 0$ and solve for $x$

To find the critical points, we set $f'(x) = 0$: $15 - \frac{5}{x^2} = 0$ Rearranging terms: $\frac{5}{x^2} = 15$ $x^2 = \frac{5}{15} = \frac{1}{3}$ $x = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$

So, the critical points are $x = \frac{\sqrt{3}}{3}$ and $x = -\frac{\sqrt{3}}{3}$.

Step 3: Determine the sign of $f'(x)$ on the intervals

To determine where $f(x)$ is increasing or decreasing, we test the sign of $f'(x)$ in the intervals determined by the critical points:

• Interval $(-\infty, -\frac{\sqrt{3}}{3})$
• Interval $(-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3})$
• Interval $(\frac{\sqrt{3}}{3}, \infty)$

Test Interval 1: $x < -\frac{\sqrt{3}}{3}$

Choose $x = -1$: $f'(-1) = 15 - \frac{5}{(-1)^2} = 15 - 5 = 10 \quad (\text{positive})$

Test Interval 2: $-\frac{\sqrt{3}}{3} < x < \frac{\sqrt{3}}{3}$

Choose $x = 0.1$: $f'(0.1) = 15 - \frac{5}{(0.1)^2} = 15 - \frac{5}{0.01} = 15 - 500 = -485 \quad (\text{negative})$

Test Interval 3: $x > \frac{\sqrt{3}}{3}$

Choose $x = 1$: $f'(1) = 15 - \frac{5}{1^2} = 15 - 5 = 10 \quad (\text{positive})$

Based on this analysis:

• Increasing on $(-\infty, -\frac{\sqrt{3}}{3}) \cup (\frac{\sqrt{3}}{3}, \infty)$
• Decreasing on $(-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3})$

Step 4: Identify Local Maxima and Minima

From the intervals where $f(x)$ is increasing or decreasing:

• The function has a local maximum at $x = -\frac{\sqrt{3}}{3}$.
• The function has a local minimum at $x = \frac{\sqrt{3}}{3}$.

Final Answers

(a) Interval(s) where $f(x)$ is increasing:
$(-\infty, -\frac{\sqrt{3}}{3}) \cup (\frac{\sqrt{3}}{3}, \infty)$

(b) Interval(s) where $f(x)$ is decreasing:
$(-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3})$

(c) x-value(s) of all local maxima:
$x = -\frac{\sqrt{3}}{3}$

(d) x-value(s) of all local minima:
$x = \frac{\sqrt{3}}{3}$

Would you like more details on any part of the solution?

Here are some follow-up questions you might consider:

1. How does the second derivative test confirm whether each critical point is a maximum or minimum?
2. What is the value of $f(x)$ at each local extremum?
3. Can you find the absolute maximum and minimum of $f(x)$ over a closed interval, such as $[-2, 2]$?
4. How does the behavior of $f(x)$ as $x \to 0$ affect the overall shape of the function?
5. How would the solution change if the function were $f(x) = 15x - \frac{5}{x}$ instead?

Tip: When finding where a function is increasing or decreasing, always look for the critical points first and use them to define intervals for testing the sign of the first derivative.