Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove: If the function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f'(x) = 0$ on $(a,b)$, then $f$ must be a constant function on $[a,b]$.

I need to select some $x_1$ and $x_2$ in $[a,b]$ such that $x_1$ is not equal to $x_2$ therefore by the mean value theorem.. This is where I am getting a bit confused how to apply MVT to this proof.. Hints/Tips appreciated. I may be over thinking things. Thanks.

share|cite|improve this question
Just a note: You should start with $x_1\in[a,b]$ fixed, and $x_2\in[a,b]$ arbitrary. Then show $f(x_2)=f(x_1)$. – David Mitra Apr 12 '12 at 17:55

$$ \begin{align} \frac{f(x_1) - f(x_2)}{x_1-x_2} & = f'(c) \text{ (for some }c\text{ between }x_1\text{ and }x_2\text{)} \\ \\ \\ & = 0 \end{align} $$ Therefore $f(x_1)-f(x_2)=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.