Does $f(x)=ax$ intersect $g(x)=\sqrt{x}$ It maybe a stupid question but I want to be sure how to explain it formally.
Does $f(x)=ax$ intersect $g(x)=\sqrt{x}$, when $x>0$ and $a>0$ (however small it is)
I think it does. 
The derivative of $f(x)$ is constant, positive. And the derivative of $g(x)$ tends to $0$. So there will be some point $x_0$, from which the derivative of $f$ will be greater than derivative of $g$. Therefore $g$ will grow slower than $f$ and both functions finally meet. Am I right? This is enough? Can one formally prove it? 

Here are the steps 
$$ f(x)=g(x) $$
$$ ax=\sqrt{x} $$
$$ (ax)^2=(\sqrt{x})^2 $$
$$ a^2x^2=|x| $$
Since $x\gt 0$ and $a\gt 0$, then
$$ a^2x^2=x $$
$$ \frac{x^2}{x}=\frac{1}{a^2} $$
Thus, when $x\gt 0$ and $a\gt 0$, the functions $f(x)$ and $g(x)$ will intersect only when $$x=\frac{1}{a^2}$$
