Determine the intervals where the graph of function is concave up and concave down and identify the inflection point when f(x)=tan^-1x?

1 Answer
Apr 13, 2018

Recall that

#d/dx(tan^-1(x)) = 1/(1 + x^2)#

Thus

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

Concavity is determined by the second derivative.

#f''(x) = (0(1 + x^2) - 2x)/(1 + x^2)^2#

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

This will have possible inflection points when #f''(x) = 0#.

#0 = 2x#

#0= x#

As you can see the sign of the second derivative changes at #x= 0# so the intervals of concavity are as follows:

#f''(x) < 0#--concave down: #(0, oo)#
#f''(x) > 0#--concave up: #(-oo, 0)#.

The graph of #f(x)# depicts.

graph{arctan [-11.25, 11.25, -5.625, 5.625]}

Hopefully this helps!