Question

Determine the values of $x$ for which the function $f(x) = \frac{x-4}{x+1}$ is an increasing or a decreasing function.

Hint:

For rational functions, the first derivative can be used to find intervals where the function is increasing or decreasing. A positive derivative indicates an increasing function.

Answer 1

We are given the function: \[ f(x) = \frac{x-4}{x+1}. \] To determine where the function is increasing or decreasing, we need to find the first derivative of $f(x)$ with respect to $x$. Using the quotient rule: \[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}, \] where $u = x-4$ and $v = x+1$. Differentiating $u$ and $v$: \[ \frac{du}{dx} = 1, \quad \frac{dv}{dx} = 1. \] Now, applying the quotient rule: \[ f'(x) = \frac{(x+1)(1) - (x-4)(1)}{(x+1)^2} = \frac{(x+1) - (x-4)}{(x+1)^2} = \frac{5}{(x+1)^2}. \] Since the derivative is always positive (as $5/(x+1)^2>0$ for all $x \neq -1$), the function is always increasing for $x \neq -1$. Thus, $f(x)$ is an increasing function for all values of $x$ except $x = -1$.