Question

Find the interval in which $f(x) = x + \frac{1}{x}$ is always increasing, $x \neq 0$.

Hint:

A function is increasing where its derivative is positive. Here, we solved for the derivative of $f(x)$ to find when it is greater than zero.

Answer 1

To determine when the function is increasing, we find the derivative of $f(x)$ and set it greater than zero. Given: \[ f(x) = x + \frac{1}{x} \] Differentiate: \[ f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2} \] Now, for the function to be increasing, we need: \[ f'(x) > 0 \] \[ 1 - \frac{1}{x^2} > 0 \] \[ \frac{1}{x^2} < 1 \] This implies: \[ x^2 > 1 \quad \Rightarrow \quad |x| > 1 \] Thus, the function is increasing for: \[ x > 1 \quad \text{or} \quad x < -1 \] Therefore, the function is always increasing on the intervals \( (-\infty, -1) \cup (1, \infty) \).