Proof $x^3+ax^2+bx+c-e^x=0$ has at least one solution

Question

I need to show that the following equation has at least one solution: $$x^3+ax^2+bx+c=e^x$$ where $a,b,c \in \mathbb{R}$.

What I did:
I defined the real-valued function $f(x)=x^3+ax^2+bx+c-e^x$ and showed that for $c=1$ we get $x=0$ as a solution, and that for $c>1$ we get a solution as $f(0)>0$ and $\lim_{x \to \infty} f(x)=- \infty$
but Im stuck with the case of $c<1$.
Any suggestions? or other ways of proving this statement. any help will be much appreciated

Answer 1

It is false, in general : note that ${x^3 \over 6}< \exp x$.

Indeed if $x<0$ this is clear as the rhs is $<0$, and for $x>0$ this follows the Taylor series of the exponential $\exp x = 1+x+ x^2/2+x^6/6+ \sum _{n>3} x^n/n!$.

Then, for all $x$, $x^3< \exp (x+ \ln 6)$.

With $y= x+\ln 6$; we get that for all $y$, $(y-\ln 6)^3 < \exp y$, a counter example

Answer 2

The result is false. Let be $f(x)=e^x-x^3+Cx^2$. It is a convex function for some $C\in\mathbb{R}$, since $f(x)''=e^x-6x+C$, let $h(x)=e^x-6x$, it has minimum on $\log(6)$, and $h(\log(6))=6(1-\log(6))$, now take $C=6\log(6)$, and you have that $f(x)''=6>0$. So that, as $f$ is strictrly convex it has at most one minimum, let $m=\inf f$, the function $g(x)=f(x)+|m|+1$ is of the form $e^x-p(x)$, where $p(x)$ is a real monic polynomial (leader coefficient equal to $1$) of degree 3, so that your result is not true in general.