there is a function which is differentiable continuously on [a,b] .

suppose

|f'(x)|<1

for all

prove that there exists

0<=k<1

that this equation is true

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- Feb 28th 2009, 09:22 AMtransgalacticdifferentiation prove..
there is a function which is differentiable continuously on [a,b] .

suppose

|f'(x)|<1

for all

prove that there exists

0<=k<1

that this equation is true

- Feb 28th 2009, 09:38 AMAbu-Khalil
If you take and the equation isn't true :S

- Mar 3rd 2009, 10:30 AMtransgalactic
i got a solution to this..but i cant understand it.

they make mean value theorem on f'(x) [x1,x2] so x2>=x1

then they say that by the second law of weirsstas there is a minimum and maximum

so there is l,L on [a,b] so

f'(L)>=f'(x)>=f'(l) for every x in [a,b]

i cant understand what is the second law of weirshtass

that gives two extreme points??

there is another things but i got stuck on this

?? - Mar 3rd 2009, 10:55 AMbkarpuz
- Mar 3rd 2009, 11:05 AMtransgalactic
what is the meaning of "I correct yours."

?? - Mar 3rd 2009, 11:12 AMbkarpuz
- Mar 3rd 2009, 11:19 AMtransgalactic
it wasnt a mistake the function is continuesly differentiable

so i can do this thing with f'(x) - Mar 3rd 2009, 04:38 PMmr fantastic
- Mar 3rd 2009, 11:47 PMtransgalactic
it have been answered there

i am trying to solve it. - Mar 4th 2009, 12:02 AMbkarpuz