$a<b$ and $f(a)=f(b)$ then $f \equiv 0$ in $[a,b]$ when $f´(x)=[f(x)]^2$ Let $f:\mathbb{R} \to \mathbb{R}$ a diff. function satisfying $f´(x)=[f(x)]^2 \;\;\forall x \in \mathbb{R}$. Show that if $a&lt;b$ and $f(a)=f(b)$ then $f$ is the null function in $[a,b]$.

I tried this: $f´(x)=[f(x)]^2\geq 0$, so $f$ is non decreasing on $[a,b]$,  which means $f(x)\geq f(a) \;\; \forall x \in [a,b]$. If there is $c \in [a,b]$ such that $f(c) \neq 0$, then, since $a&lt;c&lt;b$, there never can be the case that $f(a)=f(b)$.

Is this correct? I don´t know if I am being rigorous enough...
$f$ is increasing as you observed, so $f(a)=f(b)$ is possible only when $f$ is a constant. [ $f(b)=f(a) \leq f(x) \leq f(b)$  gives $f(x)=f(b)$]. But then $[f(x)]^{2}=f'(x)=0$ for all $x$ so $f \equiv 0$.
I think there are some arguments missing.
The following is not exactly clear in my opinion.


  If there is $c \in [a,b]$ such that $f(c) \neq 0$, then, since $a&lt;c&lt;b$, there never can be the case that $f(a)=f(b)$.


If $f(x)\neq0$, why is $a&lt;c$? Nowhere in your assumptions does it say that $f(a)=0$.

However, there is a way to fix this argument. You have to use your assumption that $f$ is non decreasing on $[a,b]$. Can you continue?