Let $f$ be a function defined on $(a,b)$. I know that
$$f'(x)=0,\ \forall x\in(a,b)\iff f\, \text{constant in}\, (a,b)$$
I know that if the domain of the function is not an interval then this result does not hold!
But my doubt is: if I have a closed and bounded interval: ex. [a,b] and in (a,b) I have $f'(x)=0$ then can I say that the function is constant in $[a,b]$?
Example: $g(x)=\arcsin{x}+\arccos{x}$. This function has a null first derivative in $(-1,1)$, to find its constant value I have thought to take $x=1$ and compute $g(1)$...but I am not sure this is right.
EDIT since the function is continuous then surely the behavior on the right and the left of $1$ must be the same, so in for $x>1$ is constant really also in $x=1$ is constant...is it right?
"if I have a closed and bounded interval: ex. [a,b] and in (a,b) I have f′(x)=0 then can I say that the function is constant in [a,b]?"
No, this is not true in general. An easy example is $f(x)=\begin{cases} 1,&x\in(a,b), \\ 0,&x=a\ \textrm{or}\ x=b. \end{cases}\ $The fact that a function is differentiable on an open interval says nothing about how it behaves outside that interval. We would need more information about the function to conclude that it is continuous on the closed interval $[a,b]$.