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#)