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
#0 = 2x#
#0= x#
As you can see the sign of the second derivative changes at
#f''(x) < 0# --concave down:#(0, oo)#
#f''(x) > 0# --concave up:#(-oo, 0)# .
The graph of
graph{arctan [-11.25, 11.25, -5.625, 5.625]}
Hopefully this helps!