I am looking at the following function on the domain $x\geq 0$:


$$F(x)=(x+a)e^{x^2}(1-\mathrm{erf}(x))-\frac{b}{\sqrt\pi},$$


where $a>0$, $0<b<1$ are parameters. From plotting this function for different values of $a$ and $b$ it seems that there is at most one root on $x\in[0,\infty)$. But how to prove it?



A problem involving the Error Function
Indeed, there is at most one root on $x\in[0,\infty)$ -- actually, for each real $a$. Consider the equivalent equation 
\begin{equation}
 r(x):=\frac{f(x)}{g(x)}=\frac{\sqrt\pi}b \tag{1}
\end{equation}
for real $x>z\_a:=0\vee(-a)$, 
where $0<b<1$ and 
\begin{equation}
 f(x):=\frac{e^{-x^2}}{x+a},\quad g(x):=1-\text{erf}(x). 
\end{equation}
Consider the "derivative ratio"
\begin{equation}
 \rho(x):=\frac{f'(x)}{g'(x)}=\frac{\sqrt{\pi } \left(1+2 a x+2 x^2\right)}{2 (x+a)^2}. 
\end{equation}
Then 
\begin{equation}
 \rho'(x)=\frac{\sqrt{\pi } \left(a^2+a x-1\right)}{(x+a)^3}, 
\end{equation}
so that, for real $x>z\_a$, we have 
$$\rho'(x)>0 \iff a>1\text{ or } \big(0 < a \le 1\ \&\ x > x\_a:=(1 - a^2)/a\big).$$ 


Now we are ready to use [l'Hospital-type rules for monotonicity](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.389.5486&rep=rep1&type=pdf). 


*Case 1:* $a\le0$, so that $z\_a=-a$. Then $\rho$ is (strictly) decreasing (on the entire interval $(z\_a,\infty)$),
whence, by Proposition 4.1 in the mentioned paper, $r$ is decreasing, and so, equation (1) has at most one root (in $(z\_a,\infty)$). In fact, in this case there is exactly one root, since $r(z\_a+)=\infty$ and $r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. 


*Case 2:* $0<a\le1$, so that $z\_a=0$. Then $\rho$ is decreasing on $(z\_a,x\_a)$ and increasing on $(x\_a,\infty)$, whence, by Proposition 4.3 in the same paper, $r$ is decreasing on $(z\_a,c\_a)$ and increasing on $(c\_a,\infty)$, for some real $c\_a\ge z\_a$. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$ on $[c\_a,\infty)$. Hence, equation (1) has at most one root in $(z\_a,\infty)$, and that possible root must be in $(z\_a,c\_a)$. 
In fact, if $0<a\le b/\sqrt\pi$, then there is exactly one root in $[0,\infty)$, since $r(0)=\frac1a\ge\frac{\sqrt\pi}b$ and $r(c\_a)<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. 
Accordingly, if $b/\sqrt\pi<a\le1$, then there is no root. 


*Case 3:* $a>1$. Then $\rho$ is increasing (on the entire interval $(z\_a,\infty)$),
whence, by the mentioned Proposition 4.1, $r$ is increasing. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. Thus, equation (1) has no roots (in $(z\_a,\infty)$) in this case. 


