Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 32

For c>0 f(x) is defined and infinitely differentiable everywhere. For c=0 f is undefined at x=0, and for c<0 f is undefined for x<=sqrt(-c) and x>+-sqrt(-c).
Find f' an f'':
f'_c(x) = (2x)/(x^2+c),
f''_c(x) = 2*(c-x^2)/(x^2+c)^2.
1. For c<0:f' is negative for xlt-sqrt(-c), f decreases. f' is positive for xgtsqrt(-c), f increases.f'' is always negative, f is concave downward on (-oo,-sqrt(-c)) and on (sqrt(-c),+oo).
2. For c=0: (f(x)=2ln|x| )f'(x)=2/x is negative for x<0, f decreases, and positive for x>0, f increases.f''(x) = -2/x^2 is always negative, f is concave downward on (-oo,0) and on (0,+oo).
3. For c>0:f' is negative for c<0, f decreases, f' is positive for x>0, f increases. At x=0 f'(x)=0 and this is a minimum.f'' has roots at x=+-sqrt(c). F'' is negative for xlt-sqrt(c) and f is concave downward,F'' is positive for x on (-sqrt(c),+sqrt(c)) and f is concave upward,F'' is negative for xgtsqrt(c) and f is concave downward.So x=+-sqrt(c) are inflection points.

The only transitional value of c is zero.Please look at the graph here: https://www.desmos.com/calculator/89vayoctfv(the green graphs are for c>0, red for c<0 and blue for c=0)