Inflection point, understanding problem I am facing problems computing the inflection point
We have $$f(x) = (2x^2-x^3)^{1/3}$$
Let's assume $f$ is defined like: $f:\Bbb{R}\to \Bbb{R}$
(As far as I read on wikipedia, it's a matter of opinion. You could also say $f: (- \infty,2]\to \Bbb{R}$)
But we assume $f:\Bbb{R}\to\Bbb{R}$
$$f'(x) = \frac{4x-3x^2}{3(2x^2-x^3)^{\frac{2}{3}}}$$
$$f''(x) = \frac{-8x^2}{9(2x^2-x^3)^{\frac{5}{3}}}$$
Both derivatives are correct, I computed it two times and finally checked it online ( e.g. WolframAlpha).
I know that for an inflection point $f''(x) = 0$
BUT: $ -8x^2 =0 \Rightarrow x = 0$
That does not work, because $f''(0)$ would not be defined, since the denominator is $0$.
What am I doing wrong?
Thank you. 
 A: I'm not a fan of it, but some teachers call the solutions to $f"(x)=0$, PPI's.  (Possible Point of Inflection.)  It's an actual inflection point only if the concavity changes at that point.  
But also, concavity can change at a point where the 2nd derivative fails to exist, which is the case in your problem.  You can factor one $x$ out of the bottom of the first derivative, and you're left with a non-removable singularity at $x=0$..  So the first derivative doesn't exist at $x=0$.  Therefore, neither does the second derivative.  
When graphing functions, I have my students collect all the points where $f'$ and  $f''$ are zero or discontinuous.  Those are the points where increasing can change to decreasing and vv, and where concave up can change to concave down.  (I don't worry about naming the points so much.)  But finally, the answer to your question is that the concavity only may change at one of these points.  In your example, it doesn't.  
A: It is neither necessary nor sufficient that $f''(c)$ exists and $f''(c) =0$ for the graph of $f$ to have an inflection point at $c$.  Consider:


*

*$f(x) = x^{1/3}$.  The graph of $f$ has an inflection point at $(0,0)$, but $f$ is not even once differentiable at $0$:

*$f(x) = x^4$.  Then $f''(0) = 0$, but $f$ has a local minimum at $0$, not an inflection point.
Instead, the definition of inflection is a point at which concavity changes.  It's harder to encode, but you can say it this way: $f$ has an inflection point at $c$ if there exist $a$ and $b$ with $a < c < b$ such that $f$ is concave up on $(a,c)$ and $f$ is concave down on $(c,b)$, or $f$ is concave down on $(a,c)$ and concave up on $(c,b)$.  
It's not hard to show that if $f''(c) = 0$ and $f'''(c) \neq 0$, then $f$ has an inflection point at $c$.  But that is not the case you have here.  Instead, you would want to show that $f''(c)$ has one sign on one side of $0$, and the other sign on the other side.
Consider that
$$
f(x) = -\frac{8x^2}{9(2x^2-x^3)^{5/3}} =  -\frac{8x^2}{9[x^2(2-x)]^{5/3}} = -\frac{8x^2}{9x^{10/3}(2-x)^{5/3}} = -\frac{8}{9x^{4/3}(2-x)^{5/3}}
$$
If $x$ is near zero, $(2-x)^{5/3}$ is positive, and $x^{4/3}$ is positive, too.  So the quotient is negative on either side of $0$, meaning $0$ is not an inflection point.
A: We can verify B. Goddard and Matthew Leingang comments with the graph of $f(x) = (2x^2-x^3)^{1/3}$ (in blue):

