Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 33

f_c(x) is defined everywhere and is infinitely differentiable.
f_c(x) = g(cx) where g(x) = x/(1+x^2).
So
f'_c(x) = g'(cx)*c = c*(1-(cx)^2)/(1+(cx)^2)^2,
f''_c(x) = c^2*g''(cx) = c^2*2(cx)*((cx)^2-3)/(1+x^2)^3.
The graphs for negative c's are the same as for positive c's except that x is replaced with -x:
f_(-c)(x) = f_c(-x).
Therefore we can consider only nonnegative c's.Also, for c=0 f_c is identically zero.
For positive c:
f'_c is positive on (-1/c,1/c) and is negative outside. So f is increasing on (-1/c,1/c) and is decreasing on (-oo,-1/c) and on (1/c,+oo). And -1/c is a local minimum, 1/c is a local maximum.
f''_c is positive on (sqrt(3)/c,+oo), negative on (0,sqrt(3)/c), positive on (-sqrt(3)/c,0) and negative on (-oo,-sqrt(3)/c). Correspondently f is concave upward on (sqrt(3)/c,+oo) and on (-sqrt(3)/c,0), and is concave downward on (-oo,-sqrt(3)/c) and on (0,sqrt(3)/c).
So 0 and +-sqrt(x)/c are the inflection points.

Please look at the graphs here: https://www.desmos.com/calculator/lbdznlp1c2
Red graphs are for negative c's, green for positive and blue for c=0.