Conjecture. An infinitely differentiable function $f:\Bbb R \to \Bbb R$ at some $x\in \Bbb R$ has all its derivatives nonzero with identical or alternative sign if and only for any polynomial $p$ the function $f(x)-p(x)$ has at most $n+1$ roots, where $n$ is the degree of $p$.
It was show in [1] that the function $f(x)=e^x$ has the property that
Property. Every polynomial $p$ of degree $n$ intersects $f$ at no more than $n+1$ points.
What is more, every infinitely differentiable function $f:\Bbb R \to \Bbb R$ satisfying
Condition. All the derivatives of $f$ are nonzero with identical or alternative sign.
has the Property, as can be seen by any applying the result in [2] to the function $g(x)\equiv f(x)-p(x)$, $g(x)\equiv -f(x)-p(x)$, $g(x)\equiv f(-x)-p(x)$, or $g(x)\equiv -f(-x)-p(x)$.
Is the Condition necessary for the Property?
I can give you a handwaivy argument why this should be a necessary condition at least for analytic functions $f$. Also w.l.o.g. $f(0) = 0$.
Lets first look at a simple case: assume $f'(x_0) = 0$ (again w.l.o.g. $x_0 = 0$) and $f'(x) > 0 ~\forall x\neq 0$ (otherwise it's even simpler). You now have to note, that $f^{(n)}(0) = 0$ too for all $n>1$ and thus at $x=0$ we have a saddlepoint meaning that locally $f$ looks like a $3$rd-degree polynomial. Indeed it looks like $ax^3$ locally, and you can thus easily construct a linear function $mx$ that interscts $f$ $3$ times. (Note: if we had not assumed $f'(x) > 0 ~\forall x\neq 0$ then we could also have an extrem point, but, as long as the condition is only broken locally, extrem points have to come in pairs, so that makes it even simpler.)
This generalizes the following way. If $n$ is the smallest natural number s.t. $f^{(n)}(x_0) = 0$ (w.l.o.g. $x_0 = 0$) and $f^{n}(x) > 0 ~\forall x\neq 0$ then around $x= 0$ $f$ looks like $p$ with \begin{align*} p(x) = ax^{n+2}+\sum_{k=0}^n a_kx^k \end{align*} You should now be able to find a polynomial of degree $n$ that intesects $f$ $(n+2)$ times.