Given: $f(x) = x + \frac{1}{x}$, and $I$ is any interval disjoint from $[-1, 1]$.

Step 1: Find the derivative

$$f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$$

Step 2: Determine the sign of $f'(x)$

For $f'(x) > 0$, we need:

$$\frac{x^2 - 1}{x^2} > 0$$

Since $x^2 > 0$ for all $x \neq 0$, the sign depends on $(x^2 - 1)$.

$f'(x) > 0$ when $x^2 - 1 > 0$, i.e., $x^2 > 1$

This gives $|x| > 1$, meaning $x > 1$ or $x < -1$.

Step 3: Analyze the interval $I$

Since $I$ is disjoint from $[-1, 1]$, every point $x \in I$ satisfies either $x < -1$ or $x > 1$.

Step 4: Apply to interval $I$

For all $x \in I$, we have $|x| > 1$, which means $f'(x) > 0$.

Step 5: Conclusion

Since $f'(x) > 0$ for all $x \in I$, the function $f$ is increasing on $I$.

By definition, $f$ is increasing on $I$ if $x_1 < x_2$ in $I$ implies $f(x_1) \leq f(x_2)$ for all $x_1, x_2 \in I$. Since the derivative is positive throughout $I$, this condition is satisfied.

∴ $f$ is increasing on $I$.