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

The provided function is $f(x) = \frac{x-4}{x+1}$. To analyze its increasing and decreasing intervals, the first derivative $f'(x)$ is calculated. Applying the quotient rule, $\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, with $u = x-4$ and $v = x+1$. The derivatives of $u$ and $v$ are $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = 1$. Substituting these into the quotient rule yields: \[ f'(x) = \frac{(x+1)(1) - (x-4)(1)}{(x+1)^2} = \frac{x+1 - x+4}{(x+1)^2} = \frac{5}{(x+1)^2}. \] As $f'(x) = \frac{5}{(x+1)^2}>0$ for all $x eq -1$, the function $f(x)$ is increasing across its domain, excluding $x = -1$.