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

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Answer 1 · 2015-04-14T16:33:43

You need to solve: #2-x/(x+2)^3 = 0# Which has no rational and only one real, solution.

If you intended to type: #f'(x) = (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 #RR - {-2}#.

#f'(-2) does not exist, but #-2# is not in the domain of #f#, so it is not a critical point.

#(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.