# on what point these function are differentiable..

• Jan 20th 2009, 11:06 AM
transgalactic
on what point these function are differentiable..
i know that if a function is differentiable on X_0
then lim [f(x+x_0)-f(x_0)]/[x-x_0] exist

[IMG]file:///C:/DOCUME%7E1/lun/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]http://img294.imageshack.us/img294/2...4365mj9.th.gif
• Jan 20th 2009, 12:50 PM
Mathstud28
Quote:

Originally Posted by transgalactic
i know that if a function is differentiable on X_0
then lim [f(x+x_0)-f(x_0)]/[x-x_0] exist

[IMG]file:///C:/DOCUME%7E1/lun/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]http://img294.imageshack.us/img294/2...4365mj9.th.gif

A) What exactly is the first one? I believe it is $\displaystyle \lfloor x\rfloor \sin^2(\pi x)$ with $\displaystyle \lfloor x\rfloor$ being the floor function.

B)$\displaystyle f:x\longmapsto\left\{\begin{array}{rcl} x^2\cos\left(\frac{\pi}{x}\right) & \mbox{if} & x\ne 0\\ 0 & \mbox{if} & x=0\end{array}\right.$

For the second one we see that the only problem is $\displaystyle f'(0)$, to ascertain its value/existence we revert back to the definitons.

\displaystyle \begin{aligned}f'(0)&=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}\\ &=\lim_{x\to 0}\frac{x^2\cos\left(\frac{\pi}{x}\right)-0}{x}\\ &=\lim_{x\to 0}x\cos\left(\frac{\pi}{x}\right)\\ &=0\end{aligned}
• Jan 21st 2009, 07:40 AM
transgalactic
regarding A:
you descibed the function correctly
how do you choose what numbers to pick
after that i just put them into the formula
and if i get a final limit then its deferentiable on that point.
how do you choose what numbers to pick?

regarding b:
how did you know to check for 0
and only for 0
??
• Jan 24th 2009, 12:52 PM
transgalactic
how to choose what points to test
??