- #1
transgalactic
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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
[tex]
f'(x)=\lim _{h->0}\frac{f(x+h)-f(h))}{h}
[/tex]
i use
|f'(x)|<1
so
[tex]
|f'(x)|=\lim _{h->0}\frac{|f(x+h)-f(h))|}{|h|}<1
[/tex]
so f(x) is continues and differentiable
so i use mvt on x1 and x2
[tex]
f'(k)=\frac{f(x1)-f(x2)}{x1-x2}
[/tex]
and i combine that with
|f'(x)|<1
[tex]
\frac{|f(x1)-f(x2)|}{|x1-x2|}<1
[/tex]
so
[tex]
|f(x1)-f(x2)| <|x1-x2|
[/tex]
what now?
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
[tex]
f'(x)=\lim _{h->0}\frac{f(x+h)-f(h))}{h}
[/tex]
i use
|f'(x)|<1
so
[tex]
|f'(x)|=\lim _{h->0}\frac{|f(x+h)-f(h))|}{|h|}<1
[/tex]
so f(x) is continues and differentiable
so i use mvt on x1 and x2
[tex]
f'(k)=\frac{f(x1)-f(x2)}{x1-x2}
[/tex]
and i combine that with
|f'(x)|<1
[tex]
\frac{|f(x1)-f(x2)|}{|x1-x2|}<1
[/tex]
so
[tex]
|f(x1)-f(x2)| <|x1-x2|
[/tex]
what now?
Last edited: