Is x^2 strictly increasing on $[0, \infty]$ I'm new here so please let me know if I'm doing something wrong.
Me and my brother are arguing since few hours on something, and we can really not figure it out, so I'm asking you some advice.
Let assume that we have a simple function $x^2$, this function is defined, continuous and derivable on the entire Real domain. 
Of course we can prove that $f'(0) = 0$.
Therefore as from my university books function should not be strictly increasing on $[0, \infty]$ (as $f'(x)$ is not always greather than 0).
But we know so only because we know the function on the entire domain and therefore we have negative and positive limits arround x=0. 
If we assume that the function is defined only on $[0, \infty]$ and here we know just that each single value of 
$$f(x) > f(x - \epsilon)$$
and therefore should logically be strictly increasing.
What are your thoughts on this?
 A: The usual definition of a strictly increasing function on a domain $D$ is that for all $x,y\in D$ with $x<y$, you have $f(x)<f(y)$. In the case of an interval $D=[0,\infty)$ and $f(x)=x^2$, you can plainly see that $f$ is strictly increasing on $D$, since $f(y)-f(x)=y^2-x^2=(y-x)(y+x)>0$ for $0\leq x<y$. 
On the other hand, $f'(x)>0$ implies $f(x)$ is increasing on $D$. In your case, you have $f'(0)=0$. However, $f'(x)=0$ does not imply $f$ is not strictly increasing. For example, take $f(x)=x^3$, so that $f'(0)=0$, yet $f(x)$ is strictly increasing. In the case of these boundary situations, $f(x)$ would not be strictly increasing if $f'(x)=0$ for all $x$ in an interval $I$
A: That's not a theorem.
It is true that if a function $f:\mathbb{R} \to \mathbb{R}$ has $f'(x)>0$ for all $x\in [a,b]$, then $f$ is strictly increasing on $[a,b]$. The proof is an application of the mean value theorem.  
However, it is not true that if a function is strictly increasing, then it must have $f'(x)>0$ for all $x$. You have provided a counterexample.  
To answer the title: yes, the function $f(x)=x^{2}$ is strictly increasing on $[0,\infty)$ 
A: Go back to the definition of a "strictly increasing" function. $f(x)$ is strictly increasing (by the definition) if whenever $x_1 > x_2$ and $x_1$ and $x_2$ are in the domain of $f$,  $f(x_1) > f(x_2)$.  Note that we said nothing at all about derivatives, or even that $f(x)$ has to have a derivative, or even that $f(x)$ needs to be continuous!
So your function $f(x)=x^2$ defined on the domain $[0,\infty)$ is strictly increasing.
A: Sounds like an argument over definitions. Many authors give slightly different definitions of things such as "strictly increasing", all of which may not be equivalent. Personally I prefer: $f$ is increasing iff $a>b \implies f(a)>f(b)$. Under this definition, $x^2$ is indeed strictly increasing on $[0, \infty)$.
A: In fact,result that you use is the consequence of mean value theorem and it requires f(x) is differentiable in (a,b) and not [a,b).
This is where you make mistake and you show f'(x)=0 but the result you use won't work if the interval is closed interval or semi closed interval
