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? 
 A: If there is a point of intersection, it will satisfy the equation $$ax = \sqrt x\implies (ax)^2 = x \iff a^2x^2 - x = 0 \iff x(a^2x - 1) = 0\;$$
Indeed, the graphs intersect, when $x = 0$ and when $a^2x-1=0 \iff x = \frac 1{a^2}$. Since we are interested in only $x\gt 0$, the point of intersection you are looking for is $$(x, f(x)) = (x, g(x))=\left(\frac 1{a^2}, g\left(\frac 1{a^2}\right)\right) = \left(\frac 1{a^2}, \frac 1a\right)$$
Note: we know that $a > 0$ and $x>0$. Hence $$g\left(\frac 1{a^2}\right) = \sqrt{\frac 1{a^2}} = \frac 1a$$ 
A: 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 $$x=\frac{1}{a^2}$$
A: writing $f(x) = g(x)$ you arrive at the equation $a^2 x^2 = x$, which gives you two solutions $x=0$ and $x=1/a^2$.
A: Let $h(x)=ax-\sqrt x$. $h$ is continuous.
Take the derivative
$$h'(x)=a-\frac1{2\sqrt x}.$$
This expression is negative then positive, showing that $h$ has a minimum. As $h(0)=0$, the value at the minimum is negative (indeed, for $x=1/4a^2$, $h(x)=-1/4a$).
And as the function then grows to plus infinity, there must be a root.
A: In this case an algebraic proof is possible; but taking it from a more general point of view, we can consider the function
$$
F(x)=ax-\sqrt{x}
$$
defined for $x\ge0$; it's continuous and $\lim_{x\to\infty}F(x)=\infty$, because
$$
\lim_{x\to\infty}(ax-\sqrt{x})=
\lim_{x\to\infty}x(a-1/\sqrt{x}).
$$
Moreover $F(0)=0$. The derivative is
$$
F'(x)=a-\frac{1}{2\sqrt{x}}=\frac{2a\sqrt{x}-1}{2\sqrt{x}}
$$
which is positive for $\sqrt{x}>1/(2a)$ and negative for $\sqrt{x}<1/(2a)$. Thus $1/(4a^2)$ is the point where $F$ has an absolute minimum.
Since
$$
F(1/(4a^2))=a\frac{1}{4a^2}-\frac{1}{2a}=-\frac{1}{4a}<0
$$
the function $F$ assumes the value zero in one and only one point $x_0>1/(4a^2)$.

Just to make an example in which the technique above would be needed is when we consider
$$
f(x)=ax,\qquad g(x)=\log(1+x)
$$
in $[0,\infty)$ (again with $a>0$).
The situation is the same as above for $F(x)=f(x)-g(x)$ with respect to continuity and limits; also $f(0)=g(0)=0$. But now
$$
F'(x)=a-\frac{1}{1+x}
$$
and so a minimum is at $x=(1-a)/a$, provided $0<a<1$. If $a\ge1$, then $F'(x)>0$ for all $x>0$, so $F$ is increasing.
If $0<a<1$, the minimum is obviously negative (because $F(0)=0$), so there will be another root of $ax=\log(1+x)$ in the interval $((a-1)/a,\infty)$.
