The given function is $f(x) = x + \frac{1}{x}$. We are tasked with finding the set of values for which $f(x)$ increases. To do this, we need to:

1. Differentiate $f(x)$ to find its derivative, $f'(x)$.
2. Determine where $f'(x) > 0$, which will give us the interval(s) where $f(x)$ is increasing.

Step 1: Find the derivative of $f(x)$

The function is: $f(x) = x + \frac{1}{x}$ Differentiating this with respect to $x$: $f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2}$

Step 2: Solve $f'(x) > 0$

Now we solve for when $f'(x) > 0$: $1 - \frac{1}{x^2} > 0$ Rearranging the inequality: $\frac{1}{x^2} < 1$ Taking the reciprocal and square-rooting both sides: $|x| > 1$ This means $x > 1$ or $x < -1$.

Conclusion

The function $f(x) = x + \frac{1}{x}$ is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.

Would you like further details or have any questions?

Related Questions:

1. How do we find where a function decreases?
2. What happens at the points $x = 1$ and $x = -1$?
3. How do you determine whether a critical point is a local minimum or maximum?
4. What is the role of concavity in determining where a function is increasing or decreasing?
5. How can we use the second derivative to understand the behavior of $f(x)$?

Tip: Always check the sign of the derivative in different intervals to determine where the function increases or decreases.