Mean value theorem question - proving F constant - answer check

Question

The question:
Given $f$ continuity at $[a,b]$ and derivative at $(a,b)$.
It is known $f'(x)=0$ for each x belongs to $(a,b)$.
prove $f$ is constant.
My Answer: Need to prove $f(x)=k$ for each X belongs to $[a,b]$
Creating a new function: $$g(x)=f(x)-k$$
$g(x)$ is also continuity and derivative at $[a,b]$ and thus I can use the Mean value theorem
I will get $$g(b)-g(a)=0$$ because it is known the derivative equals zero, If I do derivative on $g(x)$, I will get:
$$g'(x)=f'(x)$$ $$g'(x)=0$$ Thus $$g(b)=g(a)$$
Then I just use the value $a$ and $b$ and do: $$g(b)=f(b)-k$$ $$g(a)=f(a)-k$$ and get $$f(b)=f(a)$$ and that's it Sorry for the bad english... my english is not that great, I hope it will be forgiven :(

Answer 1

Try starting here instead of defining an auxiliary function. Suppose $a\leq x\leq b$. By the mean value theorem there is a $c_x\in [a,x]$ such that $f'(c_x) = \frac{f(x)-f(a)}{x-a}$.