Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The form I learned as an undergraduate was the one that says that (under appropriate conditions) for any points a and b, there is a point c such that the derivative at c is equal to the slope of the secant determined by a and b. (This of course makes it easy to prove the zero-derivative theorem.) However, in googling for the Intermediate Value Theorem for Derivatives, I find the more literal one that says that if w is between the derivative at a and the derivative at b, then there is a point c such that the the derivative of c is w.

So, are there in fact two widely-accepted forms of the Intermediate Value Theorem for Derivatives? – and if so, are they equivalent? Of course, they would be trivially equivalent in that they are both true (because they have the same truth-value, which is the truth-table definition of equivalence), but I mean, is one easily manipulated into the other?

share|improve this question
1  
The second one is known as Darboux's theorem: en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29 –  Hans Lundmark Sep 13 '11 at 8:28
2  
And the first is the mean value theorem. I've not heard it called the intermediate value theorem for derivatives before. –  Chris Eagle Sep 13 '11 at 9:58
    
Thanks guys. It looks like I just was confused about the proper theorem titles. Case closed. –  Mike Jones Sep 13 '11 at 11:16
1  
The ordinary IVT says that if $f$ is continuous on $[a,b]$, then for any $y$ between $f(a)$ and $f(b)$, there is an $x$ between $a$ and $b$ such that $f(x)=y$. The Darboux Theorem says that if $f$ is differentiable between $a$ and $b$, then for any $y$ between $f'(a)$ and $f'(b)$, there is an $x$ between $a$ and $b$ such that $f'(x)=y$. The important thing here is that $f'$ need not be continuous! Roughly speaking, this means that if $f'$ exists on in interval but is not continuous, it is not discontinuous in the boring blowing up sense, but in the desperately wiggling sense. –  André Nicolas Sep 13 '11 at 20:12

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.