Will $ -a e^{2x} + b e^{x} - cx + d$ always have a root for positive $a,c$?

Question

Consider the following exponential polynomial $$p(x) = -a e^{2x} + b e^{x} - cx + d,$$

with $a>0,c>0$ and $b,d$ arbtirary. My question is, how could one check whether this always has a root regardless of the particular choice of $b,d$?

I did some plots for different settings and it always turned out to have a root, so I am suspecting it to be true.

My ideas so far:

• Use the intermediate value theorem. The problem here is that I am not able to manipulate $p(x)$ to get an idea on where positive and negative values should lie, depending on the paremeters.
• To gain more insight, one could do a transformation $z= e^x$, such that one is left with a "polynomial" $-az^2 + bz - c \ln z + d$. If one ignores the $\ln$ part, under some conditions on the parameters, the remaining polynomial will have one root (satisfying $z>0$). At this point at least, one can then check the sign of $p$, which will depend on the parameters again.
• Check derivatives for Minima/Maxima: Here at least one gets rid of the linear term. Still, one will have to solve something like a weighted $\cosh$, i.e. $2a e^x + ce^{-x} = b$

Thanks for any suggestions!

Answer 1

Hint: Check the limits $$\lim_{x\to \pm \infty} p(x)$$

Answer 2

$$f(-\infty)>0,f(\infty)<0$$ and the function is continous so there it always a real root.