# How do you find all relative extrema for f(x) = 8/(x^2+2)?

Oct 11, 2015

See explanation section, below.

#### Explanation:

$f \left(x\right) = \frac{8}{{x}^{2} + 2}$

$f ' \left(x\right) = \frac{- 16 x}{{x}^{2} + 2} ^ 2$

$f ' \left(x\right)$ is never undefined and is $0$ only at $x = 0$.

So the only critical number is $0$.

The denominator of $f '$ is always positive, so the sign of $f '$ is the same as that of $- 16 x$, which is simply the opposite of the sign of $x$.

On $\left(- \infty , 0\right)$, $f ' \left(x\right) > 0$ and on $\left(0. \infty\right)$, $f ' \left(x\right) < 0$.

Therefore, $f \left(0\right) = 4$ is a local maximum.

There is a local maximum of $4$ (at $0$)