Post by: TrueTears on August 02, 2009, 07:09:33 pm
find the domain of
What's your method?
Post by: kurrymuncher on August 02, 2009, 07:30:42 pm
yeah, so I guess it should be all real numbers except for -2
thats how I would do it, but Kamil probably has some pr0 method.
EDIT: I dont think its necessary to draw the gradient function. As long as you can see the cusp at -2, then you'll realise that the gradient is undefined at that point.
Post by: TrueTears on August 02, 2009, 07:42:09 pm
hmmm, just draw the original function without the modulus, then reflect everything above the x axis. then draw the gradient function, take the domain and exclude all the cusps and stuff.Yeah exactly what I did kurrymuncher, thanks.
yeah, so I guess it should be all real numbers except for -2
thats how I would do it, but Kamil probably has some pr0 method.
EDIT: I dont think its necessary to draw the gradient function. As long as you can see the cusp at -2, then you'll realise that the gradient is undefined at that point.
But yeah I was wondering if there is any rigorous algebraic non graphic method lol.
Post by: TrueTears on August 02, 2009, 07:53:19 pm
Post by: kurrymuncher on August 02, 2009, 07:54:55 pm
In general... if the curve does not have any domain restrictions or asymptotes but does have cusps, then is the domain of the gradient function R \ {cusps} ?
Yep
Post by: TrueTears on August 02, 2009, 07:55:45 pm
cool story bro (Y)In general... if the curve does not have any domain restrictions or asymptotes but does have cusps, then is the domain of the gradient function R \ {cusps} ?
Yep
Post by: /0 on August 02, 2009, 07:58:49 pm
For a function to be differentiable we require that
So
Post by: TrueTears on August 02, 2009, 08:03:19 pm