theorem about inflection points in (polynomial times exponential) We know that an exponential function "kills" a polynomial, sooner or later. Here I think of polynomials that are increasing and exponential functions that are decreasing.
For instance: $x^2e^{-x^2}$ will go to zero as $x$ grows, as will $x^7e^{-x^2}$, $(C_1\cdot x^9+C_2\cdot x^{111})e^{-x^7-3x^9}$ or any similar such function.
These functions all decrease, for large enough $x$, and go to zero, asymptotically
(think of $C_1$, $C_2$ as positive). The way I set up and think about this problem, the exponential function can have no linear term in $x$. Furthermore, the polynomial should only has positive terms, and the exponential only negative terms.
Now, my question is if there is a theorem saying that, after having reached its rightmost stationary point, and as x grows further, the function has only one inflection point, and changes exactly once from concave to convex, as it goes to zero? Is this an implication of the fact that the exponential function ultimately dominates the polynomial function?
I would very much appreciate any insight on the generality of this "hypothesis".
Thank you.
 A: Let $f(x)=p(x)e^{q(x)}$, where 
$$p(x)=p_0+p_1x+\dots+p_nx^n,\ p_n\neq0,\ n\geq1$$
$$q(x)=q_0+q_1x+\dots+q_mx^m,\ q_m\neq0,\ m\geq2$$
and there exist an $A>0$  such that, for every $x>A$:
$$p'(x)>0\mbox{ and }q'(x)<0$$
so as to ensure the appropriate monotonicity, as requested (which implies that $p_n>0$ and $q_m<0$).
Now, let us calculate $f'$:
$$f'(x)=\left(p(x)e^{q(x)}\right)'=p'(x)e^{q(x)}+p(x)q'(x)e^{q(x)}=\left(p'(x)+p(x)q'(x)\right)e^{q(x)}$$
Since $p'(x)>0$, $p$ is strictly increasing for every $x>A$. So $\lim\limits_{x\to+\infty}p(x)=+\infty$ (this is trivial) and, hence, we have that there exist a $B>0$ such that $p(x)>0$ for every $x>B$. Now, let $M=\max\{A,B\}$. Then, for every $x>M$ we have that $p(x)$ is strictly positive and increasing. 
In the same way, since $q'(x)<0$, $q$ is strictly decreasing for every $x>A$. So $\lim\limits_{x\to+\infty}q(x)=-\infty$ (this is, again, trivial) and, hence, there exist a $C>0$ such that, for every $x>C$, $q(x)<0$. Now, let $N=\max\{A,C\}$. Then, for every $x>N$, we have that $q(x)$ is strictly negative and decreasing.
Since $m\geq2\Rightarrow2n<nm+1\Leftrightarrow n-1<n(m-1)$, we have:
$$\deg(p'+pq')=\max\{n-1,n(m-1)\}=n(m-1)$$
So, the coefficient of $x^{n(m-1)}$ is $np_nq_m<0$ - this is evident if we expand $p(x)q'(x)$ - and, hence $p'(x)+p(x)q'(x)<0\Rightarrow f'(x)<0$ which means that $f(x)$ is finally strictly decreasing. Let us now set:
$$L=\inf\{y>0:p'(x)+p(x)q'(x)<0\mbox{ for every }x>y\}$$
Let us also suppose that $p'(x)+p(x)q'(x)$ has at least one real root (to ensure that the peak, mentioned in the question does exist). Then it is obvious that $f'(L)=0$. We will now calculate $f''$,noticing that $f'(x)=s(x)e^{q(x)}$, where $s(x)=p'(x)+p(x)q'(x)$ which has degree $\deg(s(x))=n(m-1)$ and the coefficient of $x^{n(m-1)}$ is $mp_nq_m<0$:
$$f''(x)=(f'(x))'=\dots=\left(s'(x)+s(x)q'(x)\right)e^{q(x)}$$
So, we have:
$$\deg(s'+sq')=\max\{n(m-1)-1,n(m-1)^2\}=n(m-1)^2$$
So, the coefficient of the maximum power of $x$ is $m^2p_nq_m^2>0$ since $p_n>0$, so, finally, $s'(x)+s(x)q'(x)>0$ and strictly increasing and, hence, $f'$ is strictly increasing and $f$ is, finally, convex.
Lastly, it is easy to prove that $$\lim_{x\to+\infty}f(x)=0$$ (by induction on the degree of $p(x)$ using L' Hospital's rule is a way).
As for the number of inflection points after $L$ (remember, $L$ has been defined to be the largest root of $f'$ and hence, $f$'s largest maximum position). Well, this is for sure non-trivial to prove or disprove.
After some - well, it's been more than a couple of minutes - experimenting, I came up with this interesting example, where $f(x)=(x^2-x+1)e^{-x^3}$:

where the green line is $f$, the blue line is $f'$ and the orange line is $f''$. The key to this is that $f'$ has no real roots (altering this a little can give us an example of $f'$ with a single root). Note that $f''$ has two real roots which are both inflection points for $f$. In general, note also that, since $\deg(f')=n(m-1)$ and $\deg(f'')=n(m-1)^2$ - where with $\deg(f)$ we mean the degree of the polynomial part of $f$ - as $m$ grows, $f''$ may have many more roots than $f'$, so it is "possible" - at least intuitively - to have more inflection points after $f$'s maximum value.
Edit: (Marching drum sound) The so-wanted counter-example is, as expected, similar to the previous, with a little tweaking; $f(x)=(x^2-x+1)e^{-x^2}$ (the colours remain the same):

Important note! What does actually matter is that $q_m$ is strictly negative. Should $p_n$ be negative, the only change would be the way the convergence of $f$ to zero is achieved. In that case, $f$ would converge to zero and being finally, "below" it - meaning it would be strictly increasing and concave.
A: The claim is cleary wrong for general functions bcz of following simple reason !
Let  $f$ be a function satisfying the assumption in question i.e " after having reached its peak, and as x grows further, the function has only one inflection point, and changes exactly once from concave to convex"
Then for any $c \in R$ the function $f+c$ also satisfies the assumptions of question. So $f$ need not to tend to zero.
However under this assumption one can prove that $$\lim_{x \to + \infty}  f(x) = c \in \Bbb R \cup \{ - \infty \} $$
Now for the case  $f(x) = p(x) e^q(x)$  where $p,q$ are polynomials with $q$ is non-constant and its leading coefficient is negative then we have 
$$  \lim_{x \to + \infty } f(x) = \lim_{x \to + \infty } p(x) e^q(x) =0 $$
And Why?  because first note that for all $ n \in \Bbb N $  $$\lim_{x \to + \infty } x^n  e^{-x} = \lim_{x \to + \infty } \dfrac{x^n}{e^{x}}= n \lim_{x \to + \infty } \dfrac{x^{n-1}}{e^{x}}= . . . = n !\lim_{x \to + \infty } \dfrac{1}{e^{x}}= 0  $$
Now Let $q(x) = - a x^m + h(x) =-\frac{a}{2}x^m -x^m  (\frac{a}{2}-  \frac{h(x)}{x^m} )$  where $\deg(h) \leq m-1$ thus
$$p(x)  e^{q(x)} = p(x) e^{-\frac{a}{2}x^m }    \times e^{-x^m  (\frac{a}{2}-  \frac{h(x)}{x^m} )}  $$
Obviously RHS$ \to  0 $.
