Just because does not imply that

Since is continuous, let for some in .

Apply the MVT to the intervals and . Let and be the corresponding points and you should see that

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- Nov 22nd 2007, 08:10 AM #1

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## Mvt

Here is an outline of my question:

-f is continuous on [a,b]

-f is differentiable on (a,b)

-f(a) = a and f(b) = b

-then there are two points c1 and c2 such that 1/f'(c1) + 1/f'(c2) = 2. I know how to do the problem f'(c1) = f'(c2) = (b-a)/(b-a) = 1. I just don't get this. How do we know that the second point c2 exists with this condition? Doesn't the Mean Value Theorem only guarantee one such point? How do we know they are distinct?

- Nov 22nd 2007, 10:07 AM #2

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- Nov 22nd 2007, 11:12 AM #3

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