Spivak calculus, Sum of a polynomial's all derivatives, chapter 11 question 26

Question

The question asked to prove if $f$ is a polynomial function of degree $n$, and $f\geq0$, then $f+f'+f''+...+f^{(n)}\geq0$

The answer given is: The minimum of $f+f'+f''+...+f^{(n)}$ occurs at point where $0=f'+f''+...+f^{(n)}$[the next term vanishing, since $f$ has degree $n$]. Since $f\geq0$,this means that the minimum occurs at a point where $f+f'+f''+...+f^{(n)}\geq0$

The question is so short, so is the answer given. Why the minimum occurs at $0=f'+f''+...+f^{(n)}$? Does this point always exist?

Answer 1

Any non-constant polynomial goes to $ \infty$ in absolute value as $x \to \pm \infty$. Hence the minimum is attained at some finite point. If a differentiable function $g$ has a minimum at $x$ then $g'(x)=0$. Take $g=f+f'+...+f^{(n)}$.

Answer 2

Yes, it exists. Since $f\geqslant0$, the degree of $f$ must be even and therefore $f$ has a minimum in $\mathbb R$.