Clarification about inflection points

Question

Quoting from wikipedia inflection point,

If the second derivative, $f''(x)$ exists at $x_0$, and $x_0$ is an inflection point for $f$, then $f''(x_0) = 0$, but this condition is not sufficient for having a point of inflection, even if derivatives of any order exist. In this case, one also needs the lowest-order (above the second) non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection, but an undulation point.

I want to understand if I got this right.

Suppose we have $f(x) = x^4$. Then

$$f'(x) = 4x^3 \qquad f''(x) = 12x^2 \qquad f'''(x) = 24x \qquad f''''(x) = 24$$

If I understood well, $x = 0$ is not an inflection point for $x^4$ since the lowest nonzero derivative, above the second, is the fourth-derivative, which is of even order.

If instead I had $f(x) = x^5$, then $x = 0$ would be an inflection point.

Am I correct?

Thank you. Sorry, I just wanted a naive feedback about this.

Answer 1

You are right, the function $f(x)=x^4$ is convex on $\mathbb R$, so $x=0$ is an undulation point of $f$. In contrast, the function $g(x)=x^5$ is convex on $[0,\infty)$ and concave on $(-\infty,0]$, so $x=0$ is an inflexion point of $g$.