Show that $e^{x}=\left | P(x) \right |$ has at least one real solution

Question

We have $P$ a polynomial that is not identically $0$. Show that $e^{x}=\left | P(x) \right |$ has at least one real solution.

I got $f(x) = e^{x}- \left | P(x) \right |$, which is continuous. I need to get an $x$ where $f$ is negative, and one more where $f$ is positive. In that case there will be an $x_{0}$ and $f(x_{0})=0$. The limit when $ x\rightarrow -\infty $ is $\displaystyle{ \lim_{x\rightarrow -\infty}f(x)=\lim_{x\rightarrow -\infty}e^{x}-\left | P(x) \right |= 0 - (+\infty) = - \infty . }$ The limit when $ x\rightarrow +\infty $ is $\displaystyle{ \lim_{x\rightarrow +\infty}f(x)=\lim_{x\rightarrow +\infty}e^{x}-\left | P(x) \right |= \lim_{x\rightarrow +\infty}e^{x}(1 - \frac{\left | P(x) \right |}{e^{x}}) }$ και το $ \lim_{x\rightarrow +\infty} \frac{\left | P(x) \right |}{e^{x}} $ είναι $\frac{+\infty}{+\infty}$ , so we use del'Hopital $n$-times (depending on $P$) and we will find that $ \lim_{x\rightarrow +\infty} \frac{\left | P(x) \right |}{e^{x}}=0 $ so $\lim_{x\rightarrow +\infty}e^{x}(1 - \frac{\left | P(x) \right |}{e^{x}})= +\infty$. We have that $f$ is continuous,so there is necessarily an $x_{0}$ where $f(x_{0})=0$. So there is necessarily at least one real solution. The problem is that $\left | P(x) \right |$ is not always differentiable. How can I fix it? Is there another solution? Thanks

Answer 1

$P$ being a polynomial then at infinity $|P(x)|\sim |a_n\,x^n|$ with $n=\deg(P)$.

Assuming the equation has no real solution, then $\begin{cases}\lim\limits_{x\to-\infty} |P(x)|=+\infty\\\lim\limits_{x\to-\infty} e^x=0\end{cases}$ implies that $\forall x\in\mathbb R,\ |P(x)|>e^x$.

Now at $+\infty$ we get $|P(x)|\sim |a_n|x^n=o(e^x)$ so we can never realize $\dfrac{|P(x)|}{e^x}>1$ since this quantity tends to zero.

So there must exists at least one real solution to our equation.

Answer 2

Perhaps the most important inequality about the exponential function is $$\tag1 e^t\ge 1+t\qquad t\in\Bbb R.$$

Let $$P(x)=a_nx^n+\ldots + a_0$$ with $a_n\ne 0$. Let $h=\frac1{|a_n|}\max\{|a_{n-1}|,\ldots, |a_0|\}\ge0$. Then for $ |x|>3h$, $$\frac{|P(x)|}{|a_nx^n|}= \left|1+\frac {a_{n-1}}{a_nx}+\ldots+\frac{a_0}{a_nx^{n-1}}\right| \begin{cases}\ge 1-\frac13-\frac19-\ldots-\frac1{3^{n-1}}>\frac12\\\le 1+\frac13+\frac19+\ldots+\frac1{3^{n-1}}<\frac32\end{cases}$$ Thus for $x>\max\left\{\frac32|a_n|(n+1)^{n+1},3h\right\}$, we have (using $(1)$ and $x>0$) $$ e^x=\left(e^{x/(n+1)}\right)^{n+1}\ge\left(1+\frac x{n+1}\right)^{n+1}>\frac {x^{n+1}}{(n+1)^{n+1}}>\frac32\left|a_n\right|x^n>|P(x)|.$$ Likewise, for $x<-\max\left\{3h,\sqrt[n]{\frac 2{|a_n|}}\right\}$, we have $$e^x<1<\frac12|a_nx^n|<|P(x)|.$$ So the continuous function $\Bbb R\to\Bbb R$, $x\mapsto e^x-|P(x)|$ takes positive and negative values. By the intermediate value theorem, it also has at least one zero.