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?
Best Answer
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 $I$