Extremal values of $(x+3)^\frac{1}{3} - x^\frac{1}{3}$ $f(x) = (x+3)^\frac{1}{3} - x^\frac{1}{3}$
I am trying to find the extremal values of $f$. I start by differentiating, equating the derivative to 0, and solving that for x:
$f'(x)= \frac{1}{3} ((x+3)^\frac{-2}{3} - x^\frac{-2}{3}) = 0$
$(x+3)^2=x^2$
$x^2 + 6x + 9 = x^2$
$x = \frac{-3}{2}$
The extremal value of $f$ is at $f(\frac{-3}{2})$.
$f(\frac{-3}{2})= 2.289428485$ (decimal approximation)
However, when I try to verify this result with a CAS, WolframAlpha says something else, and SymboLab talks about the saddle of some interval, which I am not familiar with.
Is my working correct?

I've got it.
$f''(\frac{-3}{2}) < 0$, therefore the extremal is a maximum point.
