First we take the derivative. By some mildly tricky rewriting, we can factor this formula. Now that the derivative is nicely factored, we'll do the rest of the job. f ' is undefined when x = 0, since is undefined. Since f (0) is defined at 0, x = 0 is a critical point. Finally we want to find the roots of the derivative. We have f '(x) = x^{-1/3}(2 + 5x). The factor x^{-1/3} is never 0. The factor (2 + 5x) is 0 when , so we've found another critical point. The final answer: there are critical points at x = 0 and at . If we zoom in enough on the graph, this makes sense. The derivative doesn't exist at 0, and there appears to be a horizontal tangent line at : |