How do you find points of inflection and determine the intervals of concavity given #y=tan^-1x#?

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Answer 1 · 2017-02-21T00:43:06

See below.

#y = tan^-1 x#

#y' = 1/(1+x^2)#

#y'' = (-2x)/(1+x^2)^2#

Th denominator is always positive,

so the sign of #y''# is opposite the sign of #x#.

#y''# is positive for #x < 0# and negative for #x > 0#

The graph is concave up on #(-oo,0)# and concave down on #(0,oo)#

The inflection point is #(0,0)#.