Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 29

We can factor the function to get f(x)=\sqrt{x^2(x^2+c)}=|x|\sqrt{x^2+c}. There are several cases.

When c=0 , the domain is all real numbers, and the function becomes the simple quadratic f(x)=x^2 . This has a minimum at x=0 , and no inflection points, since the function is concave up for all real numbers.
The graph is given by

When c>0 , the function is more complicated. The domain is still all real numbers, but now the derivative is found to be (by using the original, unfactored function):
f'(x)=\frac{1}{2}(x^4+cx^2)^{-1/2}(4x^3+2cx) factor numerator and denominator
=\frac{x(2x^2+c)}{|x|\sqrt{x^2+c}} replace |x| with its definition
=\frac{2x^2+c}{\sqrt{x^2+c}}
The graph has a minimum at x=0 which we can see from the original function since it vanishes there and must be positive everywhere else.
The second derivative is f''(x)=+-\frac{x(2x^2+3c)}{(x^2+c)^{3/2}} . A sample graph is given by

When c=-k<0 , we can carry out exactly the same computations as in the positive case, except that now there are x-intercepts at x=+-\sqrt k . But the domain is only when the argument is positive, which means that x<=-k and x>=k .
The first derivative is f'(x)=\frac{2x^2-k}{\sqrt{x^2-k}} , which indicates the minimums at the x-intercepts.
Finally, the second derivative is f''(x)=+-\frac{x(2x^2-3k)}{(x^2-k)^{3/2}} which has inflection points at x=+-\sqrt{{3k}/2} .