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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: TrueTears on September 11, 2009, 10:11:15 pm

Title: Modulus
Post by: TrueTears on September 11, 2009, 10:11:15 pm

|x+3| => |2x-4|

Solve the inequality for x. (algebraically)

EDIT: nvm thanks d0minicz!

Title: Re: Modulus
Post by: Ilovemathsmeth on September 11, 2009, 10:54:53 pm

As far as I remember, it's always good to separate modulus functions. Then determine the individual component functions' domain.

|2x - 4| = 2x - 4, x> 2 ; 4 - 2x, x< 2

|x + 3| = x + 3, x >-3; - x - 3, x< -3

Thus x + 3 = 2x - 4, 4 - 2x

Thus - x - 3 = 2x - 4, 4 - 2x

Now determine overlapping domains. That is, x> -3 and x< 2 overlap. Also, x> -3 and x > 2 overlap.

Solve those equations independently, i.e. x + 3 = 4 - 2x; x = (1/3)

Then x + 3 = 2x - 4; x = 7

Substitute into respective function |x + 3| or |2x - 4| to obtain y coordinate if required.

Hope that helps! =)

Title: Re: Modulus
Post by: TrueTears on September 11, 2009, 11:20:33 pm

Thank you!!!

EDIT: did you mean x<-3 and x>2 'overlap'?

Title: Re: Modulus
Post by: TrueTears on September 12, 2009, 12:31:23 am

|x+3| => |2x-4|

First case:
x+3 > 0 , x> -3

x+3 => |2x-4|

second case:
x+3< 0 , x<-3

-(x+3) => |2x-4|

x+3 <= -|2x-4|

For the first case :

if 2x-4 > 0 , x > 2

x+3 => 2x-4

x<=7

if 2x-4< 0 , x<2

x+3 => -(2x-4)

x+3 => -2x+4

3x => 1

x => 1/3

Thus the answer for first case is 1/3<=x<=7

For the second case:

if 2x-4 >0

x+3 <= -2x+4

3x <= 1

x<=1/3

If 2x-4 <0

x+3 <= 2x-4

x=>7

Thus there is no intersection for this case, so no solution

Overall: the answer is 1/3<=x<=7

Title: Re: Modulus
Post by: ryley on September 12, 2009, 10:44:37 am

So is that way the preferred method to treating the moduli as sqrt(f(x)^2), squaring both sides, expanding both sides and solving from there? Doing that, I got the same answer, but I was wondering whether or not its suitable as a general method for these.

Title: Re: Modulus
Post by: TrueTears on September 12, 2009, 01:04:05 pm

Quote from: ryley on September 12, 2009, 10:44:37 am

So is that way the preferred method to treating the moduli as sqrt(f(x)^2), squaring both sides, expanding both sides and solving from there? Doing that, I got the same answer, but I was wondering whether or not its suitable as a general method for these.

Yeah there's a few way of doing it, you could also divide both sides by a say |x+3| so then you can combine the mod into one single mod.

Title: Re: Modulus
Post by: ryley on September 12, 2009, 03:23:54 pm

But are there any drawbacks to the various methods, or situations where certain methods aren't applicable/suitable?

Title: Re: Modulus
Post by: TrueTears on September 12, 2009, 03:25:44 pm

The one I did above was the longest way, a lot of IFFs and too many different cases.

I guess the other methods are much quicker and shorter, I'd definitely use them in an exam.

Title: Re: Modulus
Post by: Mao on September 12, 2009, 07:41:33 pm

Quote from: ryley on September 12, 2009, 10:44:37 am

So is that way the preferred method to treating the moduli as sqrt(f(x)^2), squaring both sides, expanding both sides and solving from there? Doing that, I got the same answer, but I was wondering whether or not its suitable as a general method for these.

It is generally a lot easier to deal with it as piecewise functions.