Does the product of an exponential decrease with a polynomial with positive coefficients have a unique maximum? Consider a function of the form
$$f(x)=\exp(-x)p(x)$$
where $p$ is a polynomial with exclusively positive coefficients. Is it true that $f$ has a unique maximum? It seems like it when playing with functions like this using a graphing calculator. And intuitively the exponential has the longest "spoolup time" but once it starts decreasing faster than the polynomial increases it only ever gets worse.
But I can not prove it. Even nicer would be an explicit representation of this maximum. The first order condition
$$
0 = f'(x) = \exp(-x)[p'(x) - p(x)]
$$
would require $p(x) = p'(x)$. Although this does not look like it should not have a unique solution...
 A: Here are two counter-examples with $2$ local maxima on $\mathbb R^+$, that I found playing around with the coefficients, and trying to keep the polynomial as simple as possible (I focused on $x^n+ax+b$)
$p(x)=10x^3+21x+18$
$q(x)=x^5+40x+20$

The positivity constraint makes it a bit difficult to reverse engineer something like "let solve $p'-p=(x-a_1)(x-a_2)\cdots(x-a_n)$ so that the derivative has many critical points".
Note that the condition on coefficients positivity implies that $p\nearrow$ on $\mathbb R^+$ and since $p(0)\ge 0$ then except maybe in zero, $p$ has not root on $\mathbb R^+$. Therefore the oscillations we see on $p(x)e^{-x}$ results only from the compared growth of the two components and not from $p(x)$ own oscillations.
A: This is not true. Take $a \in \mathbb{R}$ and
$$
p_a(x)=(x-a)^2+2(x-a)+2=x^2+2(1-a)x+1+(1-a)%2.
$$
All the coefficients of $p_a$ are positive provided $a<1$.
Now, define
$$
f(x)=p_a(x)e^{-x}, \quad a<1.
$$
Then
$$
f' (x)=-(x-a)^2e^{-x},\quad f''(x)=[(x-a)^2-2(x-a)]e^{-x}.
$$
Thanks to the first derivative test, we know that $f$
is decreasing on $(-\infty,a)$ and on $(a,\infty)$.
Because $f''(x)<0=f''(a)=f''(2)$ for $a<x<2 $ and $f''(x)>0$ for $x<a$ or $x>2$, we know that $f$ has an inflection point at $x=a$.
Thus $f$ has no maximum at all.
A: Here is another class of functions of this form which have no maximum:
Note that
$$e^xf'(x) = (p'-p)(x) = (a_1-a_0)+(2a_2-a_1)x+ \cdots +(na_n-a_{n-1})x^{n-1}-a_nx^n.$$
So, for $f'(x') = 0$, the polynomial on the RHS must equal zero at $x'$.  If we take $a_0 = a_1$, $a_2 = a_1/2,\dots$, then $(p'-p)(x) = -a_nx^n$.
Of course, this will not equal zero unless $x' = 0$.  But $x'=0$ is not a maximum unless $n = 1$.  So, for all $n \geq 2$, no maximum exists.
