Is $f''(x)=0$ sufficient for inflection point? I'm a bit confused about $n$th derivative test.Is $f''(x)=0$ at a point sufficient to prove it is inflection point or not ?Or we need to check further if any higher odd derivative is $0$? 
And when is it sufficient to conclude that if at a point of a function $f''(x)=0$ it is an inflection point ?
Assume the derivative is 0 that that point.
 A: By definition, $x_0$ is an inflection point of $f(x)$ if the concavity of $f(x)$ switches at this point. We know that $f(x)$ is concave up at $x$ if $f''(x) > 0$ and it is concave down at $x$ if $f''(x) < 0$. Therefore $x_0$ is an inflection point of $f(x)$ exactly when the sign of $f''(x)$ changes at $x_0$.
For instance, if $f''(x) < 0$ for all $x<x_0$, $f''(x_0) = 0$, and $f''(x) > 0$ for all $x>x_0$, then $x_0$ is an inflection point. More generally, you only need to check that $f''(x) < 0$ for all $x$ which are less than but close to $x_0$, i.e. for all $x$ in the interval $(x_0 - \epsilon, x_0)$ for some small positive number $\epsilon$. You would similarly need to check that $f''(x) > 0$ for all $x$ greater than, but close to $x_0$. Or of course, you can reverse the signs, so that $f''(x)$ changes from positive to negative. You do need to check that the sign changes though, in any event.
A: No it's not sufficient, it's only necessary. For $f$ to have a inflexion point at $x$, the sign of $f''(x)$ must change at the point $x$.
