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Two functions

In other words, what is a a conventional way to know if the function $f(x)$ is greater than $g(x)$, taking into account that sometimes they exchange the highest position after intersecting.

Thus a more specific question could be: How to know on each range, which function is greater than the other?

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    With continuous functions like in the image you just need to calculate the points of intersection (algebraically; that is, in your example, to solve $x^2=\sqrt{x}$ to get $x=0,1$) and then try with some easy number within each interval. You could try $x=2$ to see that $\sqrt{x}\geq x^2$ in the interval $[0,1]$. And try $x=2$ to get $x^2\geq\sqrt{x}$ in $[1,+\infty]$. – 2017-01-22

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Compute the roots of $h(x)=f(x)-g(x)$.

Then you know what intervalls are interesting.

Those intervalls are a finite number for most excercises you encounter, or have a very easy to recognize pattern (like sin(x)-cos(x) ).

Then check with values inside those intervalls, wether f(x)>g(x) or vice versa.

Can you explain why this works?

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    Cool, that's a good way to do it. I'm using this for calculating areas of intersecting functions. – 2017-01-22
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There's only one way in general: plug in a number and see what you get. If $f(x) > g(x)$ for that value of $x$, then $f$ will be larger than $g$ until they intersect again. I assume, though, that you're looking for something simpler, something you could use when $f$ and $g$ are hard to compute. The answer is that there isn't any technique that will work all the time - there's just a few tricks you can use in specific cases.

For example, in your image you're looking at the functions $f(x) = x^2$ and $g(x) = \sqrt{x}$. When $x$ is very small (very close to zero), $x^2$ is very small (a very small piece of a very small number) while $\sqrt{x}$ is "smallish" (only its square is "very small"). So $\sqrt{x} > x^2$ when $x$ is a little bigger than zero, and it will remain that way until the curves intersect again (at $x = 1$).

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Over the same interval $[a,b]$, a function $f(x)$ is higher (or lower) than function $g(x)$ if:

$f(a)>g(a) (or <)$ and $f(x)-g(x)=0$ has no root in $(a,b]$.

Your $f(x)=x^2$ and $g(x)=\sqrt{x}$ can be compared over $[1,\infty)$ as $f(1)=g(1)$ at the initial point but $x^2-\sqrt x=0$ has at most $1$ positive and no negative real root in $\mathbb R$ (Descarte's rule) hence no roots in $(1,\infty)$. So $f(x)>g(x)$ in $(1,\infty)$.

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Observe that the domain of $\sqrt{x}$ is $x>0$ and $x=0$ and that there is no restriction on $x^2$. So the restriction on both of these domains is the common domain i.e. that of $\sqrt{x}$.

Then finding where one function is greater than the other is a matter of solving the inequality $x^2>\sqrt{x}, x>0$ which works out to $x(x^3 - 1)>0$ which can be graphed and solved.