# Thread: how to finish this question..

1. ## how to finish this question..

f is differentiable continuously(which means that f(x) and f'(x) are continues and differentiable) on [a,b]
suppose that |f'(x)|<1 for x in [a,b]

prove that there is 0<=k<1 so there is x1,x2 in [a,b]
|f(x1)-f(x2)|<=k|x1-x2|
??

i started from the data that i was given and i know that f(x) is differentiable
so
$\displaystyle f'(x)=\lim _{h->0}\frac{f(x+h)-f(h))}{h}$
i use

|f'(x)|<1
so
$\displaystyle |f'(x)|=\lim _{h->0}\frac{|f(x+h)-f(h))|}{|h|}<1$

so f(x) is continues and differentiable
so i use mvt on x1 and x2
$\displaystyle f'(k)=\frac{f(x1)-f(x2)}{x1-x2}$
and i combine that with
|f'(x)|<1

$\displaystyle \frac{|f(x1)-f(x2)|}{|x1-x2|}<1$
so
$\displaystyle |f(x1)-f(x2)| <|x1-x2|$

what now?

2. they sign L as the minimum of f'(x)
and they sign l as the maximum of f'(x)

then they say that k=max{|f(L)|,|f(l)|}

but its not true
i am given that 0=<k<1
and i dont know if i get the same values

why
k=max{|f(L)|,|f(l)|} equals 0=<k<1
??