Domain

[TrueTears]

find the domain of

What's your method?

[kurrymuncher]

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, 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.

[TrueTears]

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, 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.

Yeah exactly what I did kurrymuncher, thanks.

But yeah I was wondering if there is any rigorous algebraic non graphic method lol.

[TrueTears]

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} ?

[kurrymuncher]

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

[TrueTears]

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

cool story bro (Y)

[/0]

For a function to be differentiable we require that

So is not differentiable.

[TrueTears]

Thanks /0!