1. ## Zero Derivative Theorem

I need to prove the zero derivative theorem. The theorem states that if a differentiable function f defined on the closed interval [a,b] has for every x in (a,b) f has a zero derivative at x, then the function f is constant.
I have an intuitive understanding of why this is true, but I'm not sure how to make this formal. Any help would be great. Thanks

2. Originally Posted by ion599
I need to prove the zero derivative theorem. The theorem states that if a differentiable function f defined on the closed interval [a,b] has for every x in (a,b) f has a zero derivative at x, then the function f is constant.
I have an intuitive understanding of why this is true, but I'm not sure how to make this formal. Any help would be great. Thanks
Problem: Let $\displaystyle f$ be a differentiable function defined on $\displaystyle [a,b]$ such that $\displaystyle f'(x)=0\quad\forall x\in(a,b)$ Prove that $\displaystyle f$ is a constant function.

Proof: Let $\displaystyle (\varepsilon,\varepsilon_1)$ be some arbitrary subinterval of $\displaystyle (a,b)$. By the mean value theorem we know that there exists some $\displaystyle \delta\in(\varepsilon,\varepsilon_1)$ such that $\displaystyle \frac{f(\varepsilon_1)-f(\varepsilon)}{\varepsilon_1-\varepsilon}=f'(\delta)$ but since $\displaystyle \delta\in(a,b)$ we have by assumption that $\displaystyle f'(\delta)=0$. Therefore $\displaystyle f(\varepsilon_1)-f(\varepsilon)=0$. Since $\displaystyle \varepsilon_1,\varepsilon$ were arbitrary the conclusion follows.

Remark: In retrospect, all though I think the above is fine it may have been more intuitive to consider any arbitrary subinterval $\displaystyle (a,\varepsilon)\subset (a,b)$ which would then show that $\displaystyle f(a)=f(\varepsilon)$ for any arbitrary $\displaystyle \varepsilon\in(a,b)$

3. Ok thank you that was very helpful.