im confused

[cinnamonbun]

i was just looking over insight 2008 exam 1 and question 1bwas anti diffing 1/(x-3)
and i didnt add absolute value signs cos i was told that was only used in spesh but then the answers tell you "not to forget the signs" and u cant really do the whole question unless you have the absolute value signs

so now i am confused... help anyone? 

[bec]

i think you need the absolute value signs

[polky]

Use absolute value signs!

[trinon]

i think you need the absolute value signs

Use absolute value signs!

[cinnamonbun]

argh fine = =
so do i ALWAYS add them? or only when i need them?

[ReVeL]

Yeah, you do have to include the modulus sign.

Consider the graph with its domain of

The antiderivative will be

However, if you look at the graph of the antiderivative, is only defined for when , despite the original graph being defined for excluding .

THEREFORE, you must include the absolute signs in the function of the antiderivative.

Hence, it is which has the required domain of , (check by sketching on the calculator).

Hope that helps abit.

EDIT: haha INTERGRATING is what I meant to do.

[fredrick]

log of a negative number is undefined. So if for any x the inside is negative use the modulus other wise it doesnt matter. example:
doesnt require modulus cause for all real x.

requires modulus cause is not always >0 for some x.

[fredrick]

Yeah, you do have to include the modulus sign.

Consider the graph with its domain of
The derivative will be

However, if you look at the graph of the derivative, is only defined for when , despite the original graph being defined for excluding .

THEREFORE, you must include the absolute signs in the function of the derivative.

Hence, which has the required domain of , (check by sketching on the calculator), and allows us to find the gradient of the curve when while the first one doesn't.

Hope that helps abit.

u mean antiderivative?

[nerd]

Yeah, you do have to include the modulus sign.

Consider the graph with its domain of

The derivative will be

However, if you look at the graph of the derivative, is only defined for when , despite the original graph being defined for excluding .

THEREFORE, you must include the absolute signs in the function of the derivative.

Hence, which has the required domain of , (check by sketching on the calculator), and allows us to find the gradient of the curve when while the first one doesn't.

Hope that helps abit.

I think you've made a mistake - you should be integrating rather than differentiating!

[cinnamonbun]

log of a negative number is undefined. So if for any x the inside is negative use the modulus other wise it doesnt matter. example:
doesnt require modulus cause for all real x.

requires modulus cause is not always >0 for some x.

okie dokes
thanks everyone ^o^
i was given the impression by my peers at school and stuff that the absolute signs were just for spesh o_o

thanks revel i get wut ur saying :]

[ReVeL]

Yep - accidentally typed derivative.

Sorry. :S