Given $f^\prime(p) = f^{\prime\prime}(p) = 0, f^{\prime\prime\prime}(p)\neq 0$ prove that $p$ is not a local minimum or maximum.

Let $ f:(a,b)\to \mathbb{R}$ and $p\in (a,b)$. Assume that $f$ is $3$ times differentiable at p, with $$f^\prime(p) = f^{\prime\prime}(p) = 0, f^{\prime\prime\prime}\neq 0.$$

Prove that $p$ is not a local minimum or a local maximum point of $f$.

My attempt:

Using Taylor's theorem, $\exists c \in (a,b)$ such that:$$f(x) = f(p) + f^\prime(p)(x-p) + \frac{1}{2!}f^{\prime\prime}(p)(x-p)^2 + \frac{1}{3!}f^{\prime\prime\prime}(c)(x-c)^3.$$ Thus the expression (using the conditions provided) reduces to:$$f(x) = f(p) + \frac{1}{6}f^{\prime\prime\prime}(c)(x-c)^3.$$ If we assume that $p$ is a local minimum, then for $p$ sufficiently close (i.e. $\exists \delta$ such that for $x\in(p-\delta, p+\delta), f(p) < f(x)$ and vice-versa if $p$ is a local maximum). However, I cannot see if this line of reasoning will lead to a contradiction. Am I on the right tracks?