What does third derivative tell about inflection point? I was trying to find the nature (maxima, minima, inflection points) of the function $$\frac{x^5}{20}-\frac{x^4}{12}+5=0$$
But I faced a conceptual problem. It is given in the solution to the problem that $f''(0)=0$ and $f'''(0) \neq 0$ so $0$ is not an inflection point. But why should we check the third derivative?
Isn't checking first and second derivative sufficient for verifying an inflection point ? 
Why must the higher order odd derivatives be zero for an inflection point?
 A: An inflection point is where a curve changes from concave to convex or vice versa. There are two types of inflection points: stationary and non-stationary. Stationary means that at this point the slope (thus $f'$) is $0$. These points are also called saddle-points.
Non-stationary inflection points are different. They are where the slope is at maximum, i.e. you have to maximize $f'$ in order to find them. You know from caculus that you need to look at both the $f'$ and $f''$ derivatives to determine whether a $f$ has a maximum. But now, since we look for maxima of $f'$, we have to look at $f''$ and  $f'''$.
A: 
It is given in the solution to the problem that $f''(0)=0$ and $f'''(0) \neq 0$ so $0$ is not an inflection point.

That makes no sense.
If $f''(a)=0$ and $f'''(a) \ne0$ (i.e. $f'''(a)$ is actually some well defined number and that number is not $0$), and then there is an inflection point at $a.$
But if $f''(a)=0$ and $f'''(a)=0,$ that information does not justify such a conclusion. In some cases where this happens there is an inflection point at $a$ and in some cases there is not.
To see this, assume $f''(a)=0$ and $f'''(a)>0.$ Then there is some open interval about $a$ in which $f''$ is positive to the right of $a$ and negative to the left. That means $f''$ changes sign at $a,$ so there is an inflection point.
And similarly if $f'''(a)<0.$
But $f(x)=x^4$ has $f''(0)=0$ and $f'''(0)=0$ and there is no inflection point anywhere, whereas $f(x) = x^5$ has $f''(0)=0$ and $f'''(0)=0$ and there is an inflection point at $0.$
A: It depends on your definition of inflection point. You have given this as 
"Where curve changes its concavity".
In this case checking $f'''(x)$ is not necessary since for example,
$$f(x) = x^3 \text{ gives } f'''(x) = 6 \neq 0 $$ however $x  = 0$ is an inflection point since there is a change in concavity here.
A: The given solution appears to be partially incorrect.
In fact
\begin{align}
f''(x) &= x^3 - x^2 = (x - 1)x^2 & f''(0) &= 0 & f''(1) &= 0 \\
f'''(x) &= 3x^2 - 2x & f'''(0) &= 0& f'''(1) &= 1, \\
\end{align}
and $f''$ has a double root (and local maximum) at $x=0$ and a single root at $x=1.$
At $x=0$, $f''$ is negative both on the left side and on the nearby right side of $x=0,$ but $f''$ is negative to the left of $x=1$ and positive to right, so $f$ has an inflection point at $x=1$ and not at $x=0.$
The value of the third derivative gives useful information at $x=1$ but not so much at $x=0.$ As noted in a comment, however, the fourth derivative at $x=0$ provides some useful information.
