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

Let $f:[a,b] \rightarrow \mathbb{R}$ be a function which is continuous on $(a,b]$ and differentiable on $(a,b)$. Is there any function such that $f(b)-f(a)≠(b-a)f'(x), \forall x\in (a,b)$?

There was a typo, and now it's edited. I wanted to know whether compactness of a set where $f$ is continuous on is essential.

share|cite|improve this question
If $f \colon (a,b] \to \mathbb{R}$, then what is $f(a)$? – Antonio Vargas Nov 14 '12 at 18:28
How do you define $f(a)$ if $a$ is not in the domain of $f$? – copper.hat Nov 14 '12 at 18:28
If you think of the function $f$ as being defined for all reals, dropping the continuity at $a$ condition lets you assign $f(a)$ arbitrarily. This year, I like $f(x)=0$ on $(a,b]$ and $f(a)=2012$. – André Nicolas Nov 14 '12 at 18:32
I think this is a good way to think about theorems to see if you understand them. – Ross Millikan Nov 14 '12 at 18:35
Also, if your current version $f(b)-f(a)\ne (b-a)f'(x),\forall x\in (a,b)$ is correct, then any nonlinear function satisfies your criteria. I think you meant $\exists x\in (a,b)$. – rayradjr Nov 14 '12 at 18:36
up vote 1 down vote accepted

Let $f(0)=-20, f(x)=0 \text{ for } x \in (0,1]$. Is this the sort of example you were thinking of?

share|cite|improve this answer
Great! Thank you – Katlus Nov 14 '12 at 18:45

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.