# Interemediate Value Rule of Derivatives

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• Dec 18th 2007, 11:22 PM
Truthbetold
Interemediate Value Rule of Derivatives
If y=f(x) is continuous on [a,b], and N is a number between f(a) and f(b), then there is at least 1 c ∈ [a,b] such that f(c) = N.

That's actually the normal value. They're fundamentally the same.
So according to this, if I get it,
if y= |x| is differentiable at [-1, 1] then everything between the two is differentiable. But 0 isn't differentiable.
Something about my logic, if it's even worth calling that, is seriously weird.

Woulds someone explain my prob? Thank you!
• Dec 19th 2007, 07:09 AM
ThePerfectHacker
Quote:

Originally Posted by Truthbetold
If y=f(x) is continuous on [a,b], and N is a number between f(a) and f(b), then there is at least 1 c ∈ [a,b] such that f(c) = N.

That's actually the normal value. They're fundamentally the same.
So according to this, if I get it,
if y= |x| is differentiable at [-1, 1] then everything between the two is differentiable. But 0 isn't differentiable.
Something about my logic, if it's even worth calling that, is seriously weird.

Woulds someone explain my prob? Thank you!

A function is said to have the intermediate value property on an interval I (it can be any interval, open, closed, infinite, ... ) iff for any a<b in I we have that if N is between f(a) and f(b) in value then there exists c in [a,b] so that f(c) = N.

Now if f is differenciable on an open interval I then f' has the intermediate value property on I. That is the theorem. What you wrote does not satisfy the conditions of this theorem, which is why it is not working.