How many times does the function $y=e^x $ meet $y=x^2$? As you know $y=e^x$ and $y= x^2$ meet once on $x<0$. 
But I want to know whether or not they meet on $x>0$. 
Since $\lim_{x\rightarrow \infty } e^x/x^2=\infty$, if they meet once on $x>0$, they 
 must meet again. 
To summarize, my question is whether or not they meet on $x>0$. 
Thank you in advance. 
 A: Consider the function $f(x) = x^2 e^{-x}$, which has a maximum value of $4/e^2$ at $x=2$.  As this max value is less than $1$, this implies that $x^2$ and $e^x$ do not meet for $x>0$.
A: Note that $$e^x - x^2 = \left(e^{x/2}-x\right)\left(e^{x/2}+x\right)$$
For $x\ge 0$, the second factor is strictly positive, so consider the first factor:
$$\begin{align}e^{x/2}-x &= \left(1+\frac{x}{2}+\frac{x^2}{8} + \frac{x^3}{48} + \cdots\right) - x \\
&= 1 - \frac{x}{2}+\frac{x^2}{8} + \frac{x^3}{48}+\cdots \\[6pt]
&= \frac{1}{2} + \frac{1}{8}\left( 4-4x+x^2 \right) + \frac{x^3}{48} + \cdots \\[6pt]
&= \frac{1}{2} + \frac{1}{8} \left( 2-x \right)^2 + \frac{x^3}{48} + \cdots \\[6pt]
&\ge \frac{1}{2}
\end{align}$$
Thus, that first factor is also strictly positive, whence $e^x - x^2$ is as well. QED
A: alternatively - $e^x > x^2$ because you have equality at $0$, so you can compute the derivatives and then you need to check that $e^x > 2x$ which is true - just exapnd $e^x$ into Taylor series and use AM-GM to get $x^2 / 2 + 1 \geq x + 1/2$. you get the other $x$ straight from the expansion
A: With Lambert W
Consider the Lambert W Function, which is the inverse of $xe^x$. We can write
$$
\begin{align}
x^2e^{-x}&=1\\[6pt]
xe^{-x/2}&=\pm1\\[6pt]
-\frac x2e^{-x/2}&=\mp\frac12\\
\end{align}
$$
Therefore,
$$
x=-2\,\mathrm{W}\left(\pm\frac12\right)
$$
For $+\!\frac12$, there is one real branch of $\mathrm{W}$:
$$
-2\,\mathrm{W}\left(+\frac12\right)=-0.70346742249839165205
$$
Unfortunately, for $x\in\mathbb{R}$, the minimum of $xe^x$ is $-\frac1e$, so there is no real solution for $xe^x=-\frac12$.
However, there are many complex solutions, such as
$$
-2\,\mathrm{W}\left(-\frac12\right)=1.5880472646893787359+1.5402235010207582194\,i
$$

Without Lambert W
Since
$$
\frac{\mathrm{d}}{\mathrm{d}x}x^2e^{-x}=(2x-x^2)e^{-x}
$$
on $(-\infty,0)$, $x^2e^{-x}$ decreases monotonically from $\infty$ to $0$
on $(0,2)$, $x^2e^{-x}$ increases from $0$ to $4e^{-2}$
on $(2,+\infty)$, $x^2e^{-x}$ decreases from $4e^{-2}$ to $0$
Because $4e^{-2}\lt1$, there is no $x\ge0$ so that $x^2e^{-x}=1$.
Of course, since $\lim\limits_{x\to-\infty}x^2e^{-x}=+\infty$ and at $x=0$, $x^2e^{-x}=0$, there is one unique value $x\lt0$ for which $x^2e^{-x}=1$.
