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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: cinnamonbun on November 06, 2008, 03:39:45 pm

Title: im confused
Post by: cinnamonbun on November 06, 2008, 03:39:45 pm: 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? :(
Title: Re: im confused
Post by: bec on November 06, 2008, 03:45:52 pm: i think you need the absolute value signs
Title: Re: im confused
Post by: polky on November 06, 2008, 03:50:08 pm: Use absolute value signs!
Title: Re: im confused
Post by: trinon on November 06, 2008, 03:51:29 pm: Quote from: bec on November 06, 2008, 03:45:52 pm
i think you need the absolute value signs

Quote from: polky on November 06, 2008, 03:50:08 pm
Use absolute value signs!
Title: Re: im confused
Post by: cinnamonbun on November 06, 2008, 03:52:37 pm: argh fine = =
so do i ALWAYS add them? or only when i need them?
Title: Re: im confused
Post by: ReVeL on November 06, 2008, 03:55:41 pm: Yeah, you do have to include the modulus sign.

Consider the graph $y=\frac {1}{x-3}$ with its domain of $R/(3)$

The antiderivative will be $log_e (x-3)$

However, if you look at the graph of the antiderivative, $log_e (x-3)$ is only defined for when $x>3$ , despite the original graph being defined for $R$ excluding $3$ .

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

Hence, it is $log_e|x-3|$ which has the required domain of $R/(3)$ , (check by sketching on the calculator).

Hope that helps abit.

EDIT: haha INTERGRATING is what I meant to do.
Title: Re: im confused
Post by: fredrick on November 06, 2008, 03:57:09 pm: 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:
$log_{e}(x^2+1)$ doesnt require modulus cause $x^2+1>0$ for all real x.

$log_{e}|x+1|$ requires modulus cause $x+1$ is not always >0 for some x.
Title: Re: im confused
Post by: fredrick on November 06, 2008, 03:58:31 pm: Quote from: ReVeL on November 06, 2008, 03:55:41 pm
Yeah, you do have to include the modulus sign.

Consider the graph $y=\frac {1}{x-3}$ with its domain of $R/(3)$
The derivative will be $\frac {dy}{dx}=log_e (x-3)$

However, if you look at the graph of the derivative, $log_e (x-3)$ is only defined for when $x>3$ , despite the original graph being defined for $R$ excluding $3$ .

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

Hence, $\frac {dy}{dx}=log_e|x-3|$ which has the required domain of $R/(3)$ , (check by sketching on the calculator), and allows us to find the gradient of the curve when $x<3$ while the first one doesn't.

Hope that helps abit.
u mean antiderivative?
Title: Re: im confused
Post by: nerd on November 06, 2008, 03:59:11 pm: Quote from: ReVeL on November 06, 2008, 03:55:41 pm
Yeah, you do have to include the modulus sign.

Consider the graph $y=\frac {1}{x-3}$ with its domain of $R/(3)$

The derivative will be $\frac {dy}{dx}=log_e (x-3)$

However, if you look at the graph of the derivative, $log_e (x-3)$ is only defined for when $x>3$ , despite the original graph being defined for $R$ excluding $3$ .

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

Hence, $\frac {dy}{dx}=log_e|x-3|$ which has the required domain of $R/(3)$ , (check by sketching on the calculator), and allows us to find the gradient of the curve when $x<3$ while the first one doesn't.

Hope that helps abit.

I think you've made a mistake - you should be integrating rather than differentiating!
Title: Re: im confused
Post by: cinnamonbun on November 06, 2008, 04:01:19 pm: Quote from: fredrick on November 06, 2008, 03:57:09 pm
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:
$log_{e}(x^2+1)$ doesnt require modulus cause $x^2+1>0$ for all real x.

$log_{e}|x+1|$ requires modulus cause $x+1$ 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 :]
Title: Re: im confused
Post by: ReVeL on November 06, 2008, 04:18:13 pm: Yep - accidentally typed derivative.

Sorry. :S