Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 30

This function is defined only on [-c,c] (we can assume that c is positive) and infinitely differentiable inside the interval.
f'_c(x) = sqrt(c^2-x^2)+x*(-2x)/(2sqrt(c^2-x^2)) = (c^2-2x^2)/sqrt(c^2-x^2).
f''_c(x) = (-4x*sqrt(c^2-x^2)-(c^2-2x^2)*(-2x)/(2sqrt(c^2-x^2)))/(c^2-x^2) = (x*(2x^2-3c^2))/(c^2-x^2)^(3/2).

f' is negative for x in (-c,-c/sqrt(2)) and in (c/sqrt(2),c), f decreases there, and f' is positive for x in (-c/sqrt(2),c/sqrt(2)), f increases.
So -c/sqrt(2) is a local minimum, c/sqrt(2) is a local maximum.
About f'': 2x^2-3c^2 is negative on (-c,c) so f'' is positive on (-c,0) and is negative on (0,c). f is concave upward on (-c,0) and is concave downward on (0,c). So x=0 is an inflection point.
The general shape of the graph is the same for all c's.
Please look at the graphs here: https://www.desmos.com/calculator/4a0q7l7sec