Question: Let $f(x)=e^{-x^2}:\Bbb R\to\Bbb R$. Is there some bounded interval $I=[-a,a]\subset\Bbb R$ so that all zeros of $f^{(n)}$ (the $n$-th derivative) are contained in $I$ for all $n\in \Bbb N$?
My approach was to express $f^{(n)}$ as $P_n(x)e^{-x^2}$ for some polynomials $P_n$. We find
$$\frac{\mathrm d}{\mathrm dx}[P_n(x)e^{-x^2}]=P_n'(x)e^{-x^2}-2xP_n(x)e^{-x^2}=\underbrace{(P_n'(x)-2xP_n(x))}_{P_{n+1}(x)}e^{-x^2}.$$
Together with $P_0(x)=1$, this defines exactly the (physicis's) Hermite polynomial (thanks to gammatester for pointing this out in the comments). So the question is essentially whether all zeros of the Hermite polynomials are contained in some interval $I=[-a,a]$.
No. In fact the zeroes of the Hermite polynomials are known to be asymptotically semicircularly distributed in the interval $[-2 \sqrt{n}, 2 \sqrt{n}]$; see, for example, this paper by Kornyik and Michaeletzky.