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?
 A: Your calculus is correct and $x=-\frac{3}{2}$ is a maximum as $f''(-\frac{3}{2})<0$ as noted in the answer of @dewypeters.
But note that the first derivative $f'$ is not defined for $x=-3$ and $x=0$ that are points where the graph of the function has a vertical tangent.
The result of WolframAlpha may depend on this fact and, maybe, by the particular way in which you are write the question (that I cannot see, but you can see the answers to my question about the roots of the first derivative : A WolframAlpha error?). Anyway you can see here the graph of your function by WA.
A: Take the second derivative at -3/2.
$f''(x)= \frac{1}{3} (-(x+3)^\frac{-5}{3} + x^\frac{-5}{3}) = 0$
plug (-3/2) into that.  f''(x) = 0.  Thus your point is a saddle point.  Minima are when f'' are positive.  Maxima are where f'' is negative.  Saddle points are where f'' are 0.
Max is where the function had been raising (so f' had been positive), reaches a peak when f' is 0 and goes back down when f' is neg. So f'' is neg at this point.  Min the exact opposite.
A saddle is where function had be raising (or dropping).  f' slows to zero so the function plateaus.  But then rather than dropping (or raising) it begins to raise again.  So f' went from pos to zero back to pos.  Which means f'' was zero at this point. 
A: I've got it.
$f''(\frac{-3}{2}) < 0$, therefore the extremal is a maximum point.
