Q:

Show that a polynomial function of odd degree $\geq 3$ has at least one inflection point.

If $f(x)$ is a polynomial function of degree $\geq 3$, then $f''(x)$ is also a polynomial of odd degree, and hence always have a real root.
How we argue that concavity changes about that point?

A:

As you say, $f''(x)$ is a polynomial of odd degree. This means that either $\lim_\limits{x\to-\infty}f''(x)=-\infty$ and $\lim_\limits{x\to\infty}f''(x)=\infty$, or $\lim_\limits{x\to-\infty}f''(x)=\infty$ and $\lim_\limits{x\to\infty}f''(x)=-\infty$. And $f''(x)$ is continuous, so it must have a zero.
Being a polynomial, $f''(x)$ has only finitely many zeroes, so if $f''(a)=0$, there is an $\epsilon_a>0$ such that $f''$ has no other zeroes in $(a-\epsilon_a,a+\epsilon_a)$, and hence $f''(x)$ does not change sign on $(a-\epsilon_a,a)$ or on $(a,a+\epsilon_a)$. Say that $f''$ changes sign at $a$ if the sign of $f''(x)$ on $(a-\epsilon_a,a)$ is different from its sign on $(a,a+\epsilon_a)$. If $f''$ does not change sign at any zero, it’s not hard to see that either $f''(x)\ge 0$ for all $x\in\Bbb R$, or $f''(x)\le 0$ for all $x\in\Bbb R$, both of which are impossible. Thus, $f''$ must have a zero $a$ at which it changes sign, and $f$ then has an inflexion point at $a$.

