How do you find the critical numbers of #1/(t^2+3)#?

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Answer 1 · 2017-05-26T22:23:29

A critical number for #f(t)# is a solution of the equation:

#f'(t) = 0#

In our case:

#f'(t) = -(2t)/(t^2+3)^2#

So the only critical point of the function is for #t=0#

Since #f'(t) > 0# for #t<0# and #f'(t) < 0# for #t > 0# such point is a local maximum.

graph{1/(x^2+3) [-2.5, 2.5, -1.25, 1.25]}