# How do you find the critical points if f'(x)=2-x/(x+2)^3?

Apr 14, 2015

You need to solve: $2 - \frac{x}{x + 2} ^ 3 = 0$ Which has no rational and only one real, solution.

If you intended to type: $f ' \left(x\right) = \frac{2 - x}{x + 2} ^ 3$, the we're in better luck.

A critical number is a value in the domain of $f$ at which the derivative is either $0$ or fails to exist.

It looks as if the domain for the original $f$ was $\mathbb{R} - \left\{- 2\right\}$.

$f ' \left(- 2\right) \mathrm{do} e s \neg e \xi s t , b u t$-2$i s \neg \in t h e \mathrm{do} m a \in o f$f#, so it is not a critical point.

$\frac{2 - x}{x + 2} ^ 3 = 0$ when $2 - x = 0$ which happens at $x = 2$.

Assuming that $2$ is in the domain of $f$, it is a critical point.