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

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Answer 1 · 2015-10-11T03:27:31

See explanation section, below.

$f(x) = 8/(x^2+2)$

$f'(x) = (-16x)/(x^2+2)^2$

$f'(x)$ 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 $-16x$ , which is simply the opposite of the sign of $x$ .

On $(-oo,0)$ , $f'(x) > 0$ and on $(0.oo)$ , $f'(x) < 0$ .

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

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

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