http://i453.photobucket.com/albums/q...188/math11.jpg

Any answers to this would be greatly appreciated, bit of assessed work i'm really struggling on.

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- Nov 20th 2008, 12:03 PMpkrUrgent Help!
http://i453.photobucket.com/albums/q...188/math11.jpg

Any answers to this would be greatly appreciated, bit of assessed work i'm really struggling on. - Nov 20th 2008, 12:25 PMThePerfectHacker
The idea for the first parts is that if $\displaystyle \epsilon > 0$ and $\displaystyle x$ is a fixed point then we can take $\displaystyle \delta > 0$ small enough such that $\displaystyle |x-y|^{\alpha} < \epsilon$ for all $\displaystyle |x-y| < \delta$. Therefore, $\displaystyle |f(x)-f(y)| \leq |x-y|^{\alpha} < \epsilon$. Thus, $\displaystyle f$ is continous.

In the second part if $\displaystyle \alpha > 1$ then $\displaystyle |f(x)-f(y)| \leq |x-y|^{\alpha} $ then $\displaystyle | [f(x)-f(y)]/[x-y] | \leq |x-y|^{\alpha - 1}$. Notice the LHS has form of a derivative. Thus, what can you conclude?