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I am having a problem with this exercise. Please help.

Is it possible to find a function $f$ with a continuous derivative $f'$, such that $f'(x)>0$ and

  1. $f(0)=1,\;f(1)=0$,
  2. $f(0)=-1,\;f(1)=0$?

If yes give an example, and if not, show why

Please help

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For (1) look up Rolle's Theorem. For (2) there's many choices, one of which is a linear y=mx+b. – coffeemath Oct 17 '12 at 18:33
How can I use Rolle's theorem if we don't have anywhere f(a)=f(b) ? – Carpediem Oct 17 '12 at 18:40
How do I use it then here ? – Carpediem Oct 17 '12 at 18:49
up vote 1 down vote accepted

The answer to (1) is probably most easily obtained using the mean value theorem, which is a generalization of Rolle's theorem.

As noted in the comments, there are plenty of suitable examples for (2). In particular, a straight line passing through those two points will do.

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Thanks. It's draw. :-) – vesszabo Oct 17 '12 at 19:04
  1. By Lagrange mean -value theorem there exists $0<\tau<1$ such that $$ f'(\tau)=\frac{f(1)-f(0)}{1-0}=-1, $$ which is a contradiction. So there is no such function.
  2. (Hint). Find a linear function.
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+1 for being faster. :) – Ilmari Karonen Oct 17 '12 at 18:56

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