How to show that the square function is increasing i.e. $\forall x \geq 0; f(X)=x^{2}$?

Question

I read that this means to show $\forall x \in R x > y \implies x^{2}>y^{2}$. Whether it can be shown by $f'(x) >0$ or $f'(x) \geq0$?. Because, if its the first one then f^{'}(x) =2x = 0 at x=0 and hence square function would not be increasing. please clarify on that.

Answer 1

We can show that $f(x)=x^2$ is strictly increasing as $\frac{d}{dx}x^2=2x$. Obviously, if $a>b$ and $a,b>0$, $2(a)>2(b)$, meaning that the derivative is strictly increasing $\forall{x}>0$. This proves that $f(x)$ is strictly increasing for $x>0$.