f = x - arctan(x)
f' = 1 - 1/(1+x^2) = x^2/(1+x^2)
f'=0 only at x=0
f" = 2x/(1+x^2)^2
f"(0) = 0, so x=0 is not a min or a max, but a point of inflection.
See the graph at
http://www.wolframalpha.com/input/?i=x+-+arctan%28x%29
a). Find the critical points of the following functions on the domain or on the given interval.
b). Use a graphing utility to determine whether each critical point corresponds to a local minimum, local maximum, or neither
f(x)= x- tan^-1 x
3 answers
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1 answer