$e^x= x^2+x$ has no roots for $x>0$ I need to prove using elementary calculus that:
$$e^x=x^2+x$$
has no roots for $x>0$.
I could easily observe it graphically but how can I prove it.
Please suggest.
 A: We have $e^x > 1+x+x^2/2+x^3/6$, so we show that the right side is bigger than $x^2+x$ for positive $x$.  Let
$$f(x) = 1+x+x^2/2+x^3/6-(x^2+x) = x^3/6-x^2/2+1.$$
Find the minimum value of $f(x)$ using usual calculus methods:
$$f'(x) = x^2/2-x = 0$$
gives one positive critical point $x=2$.  $f''(x) = x-1$ and $f''(2) >0$ so the critical point is a minimum.  Since $f(2) = 1/3>0$ the minimum value of $f$ on $x>0$ is positive.
A: Fix a positive integer $n$ and define
$$
f_n(x):=e^x-1-x-x^2/2-\cdots-x^n/n!.
$$
Then $f^\prime_n(x)=e^x-1-x-\cdots-\frac{x^{n-1}}{(n-1)!}$, which is positive (by induction) on $(0,\infty)$.
Hence $f_n$ in strictly increasing on $(0,\infty)$. In particular, this gives the known inequality
$$
\forall x>0, \quad 
e^x>1+x+\cdots+\frac{x^n}{n}.
$$
Suppose that $z$ satisfies $e^z=z^2+z$. Then
$$
z^2+z=e^z=f_n(z)+1+z+\cdots+\frac{z^{n}}{n!}>f(0)+1+z+\cdots+\frac{z^n}{n!}=1+z+\cdots+\frac{z^n}{n!}
$$
for all $n$. Now pick $n=1$ and you know that $z>1$. Set $y:=z-1 \in (0,\infty)$, so that
$$
y^2+3y+2=z^2+z=e^z=e\cdot e^y.
$$
Using the above inequality with $n=2$ we obtain
$$
e\cdot e^y> e\left(1+y+\frac{y^2}{2}\right)>y^2+3y+2,
$$
where the last inequality is equivalent to
$$
y^2+2y\left(1-\frac{1}{e-2}\right)+2>0,
$$
or, equivalently,
$$
\left(y+1-\frac{1}{e-2}\right)^2+\left(2-\left(1-\frac{1}{e-2}\right)^2\right)>0
$$
However, the first square is nonnegative and the second one is positive.
