Bound number of roots of nearly polynomial map

Question

Let $P,Q\in\mathbb{R}[x]$ be polynomials of degrees $p$ and $q$, respectively, and $c\in\mathbb{R}$ with $c\neq 0$. I am interested in bounding the number of real roots of the map $$x \mapsto P(x) + e^{cx} Q(x)\tag{1}.$$

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Let $\mathbb{A}$ denote the algebraic numbers. So far, all I have is the following (trivial) case, which bounds the number of algebraic roots (including possibly complex ones):

If $P,Q\in\mathbb{A}[z]$ and $c\in\mathbb{A}$ with $c\neq 0$, the map $(1)$ has at most $\min\{p,q\}+1$ algebraic roots.

Proof: If $z\in\mathbb{A}$ with $z\neq 0$, $P(z)$ and $Q(z)$ are algebraic and $e^{cz}$ is not (Lindemann–Weierstrass theorem). Therefore, the map $(1)$ can have at most $\min\{p,q\}$ nonzero algebraic roots.

Answer 1

The map $(1)$ has at most $p+q+1$ real roots.

Proof: We proceed by induction on the degree of $P$ and use the fact that a smooth real function with at most $n$ critical points has at most $n+1$ roots.

If $\deg P=0$, the derivative of $(1)$ is $$x\mapsto e^{cx}(cQ(x)+Q^\prime(x)).$$ Trivially, this map has at most $q$ zeroes. Therefore, $(1)$ has at most $q+1$ zeroes.

Suppose the statement holds true for all polynomials $P$ with $\deg P=p-1$. Fix $P$ with $\deg P=p$. The derivative of $(1)$ is $$x\mapsto P^\prime(x)+e^{cx}(cQ(x)+Q^\prime(x)).$$ Note that this has the form $(1)$, and hence by the hypothesis, has at most $(p-1)+q+1$ roots.

The desired result follows by induction.

Answer 2

A nontrivial example that achieves $p+q+1$ roots.