Numbers of solutions $e^x=-x^2+2x+5 $ Prove that $e^x=-x^2+2x+5 $ have exactly two solutions.
Is it enoguht that Vertex of the parabola is over $y=e^x$ and arms of it looks down
 A: This is a little out of the blue, but...
Since $e^x > 0$ for all $x \in \mathbb{R}$, there can only be two points where RH intersects $e^x$ since $e^x = -x^2+2x+5 \implies x^2-2 x+e^x = 5$.
If we rewrite $x^2-2 x+e^x = 5$ as $x^2-2 x+e^x = y$, observe that its minima is at $x^2-2 x+e^x = W(\frac{e}{2}) (W(\frac{e}{2})+2)-1,\, x = 1-W(\frac{e}{2})$
where $W(n)$ denotes the Lambert W function.
As $x$ tends from $1-W(\frac{e}{2}) \to \infty$ or $1-W(\frac{e}{2}) \to -\infty$, $y\to\infty$, thus it is easy to see that $e^x = -x^2+2x+5$ has only two solutions.
A: Yes. More rigorously, we can use the Intermediate Value Theorem (IVT) to prove that at least two solutions exist. Then we can use Rolle's Theorem to prove that at most two solutions exist.

Let $f(x) = e^x + x^2 - 2x - 5$. Note that $f$ is continuous for all $x$. Furthermore, note that:
\begin{align*}
f(-10) &= e^{-10} + 100 + 20 - 5 = e^{-10} + 115 > 0 \\
f(0) &= 1 + 0 - 0 - 5 = -4 < 0 \\
f(10) &= e^{10} + 100 - 20 - 5 = e^{-10} + 75 > 0 \\
\end{align*}
Hence, by IVT, we have $f(x)=0$ for some $x \in (-10,0)$ and for some $x \in (0,10)$.

Now suppose instead that the equation $f(x)=0$ has three solutions $a<b<c$. Since $f$ is continuous and differentiable everywhere and $f(a)=f(b)=f(c)=0$, the conditions of Rolle's Theorem are satisfied. Hence, we know that for some $r\in (a,b)$ and for some $s \in (b,c)$, we have $f'(r) = f'(s) = 0$.
We will now apply Rolle's Theorem a second time to the equation $f'(x)=0$. Since the function $f'(x)=e^x + 2x - 2$ is continuous and differentiable everywhere and $f'(r) = f'(s) = 0$, the conditions of Rolle's Theorem are satisfied. This yields $f''(t)=e^t + 2=0$ for some $t \in (r,s)$. But this is impossible, since $e^t$ is nonnegative for all $t$.
