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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: red bull on January 05, 2010, 02:00:01 am

Title: Redbull's Question Thread
Post by: red bull on January 05, 2010, 02:00:01 am: Hi all! i am having some difficulty with modulus can u please help me and fully explain each step on how u get the answer. Here are my questions:

1. on a number line, illustrate each of the following sets and represent the sets using interval notation.

a) {x:|x|<3}
b) {x:|x|≥5}
c) {x:|x-2|≤1}
Title: Re: Redbull Question Thread
Post by: TrueTears on January 05, 2010, 02:06:42 am: Try split it up into cases.

Eg, $|x| = x$ if $x \ge 0$

$|x| = -x$ if $x <0$
Title: Re: Redbull Question Thread
Post by: red bull on January 05, 2010, 02:14:51 am: i still don't get it...
Title: Re: Redbull Question Thread
Post by: TrueTears on January 05, 2010, 02:16:35 am: Well that's the definition of a modulus function. You can read more about in the Essentials textbook (do you use it?)
Title: Re: Redbull Question Thread
Post by: kamil9876 on January 05, 2010, 02:18:45 am: I think if a graph is supposed to be drawn, might as well use the definition "|a| is the distance between 0 and a on the number line"
Title: Re: Redbull's Question Thread
Post by: superflya on January 05, 2010, 12:52:53 pm: I love u modulus :p
Title: Re: Redbull's Question Thread
Post by: Ilovemathsmeth on January 12, 2010, 12:28:18 am: Modulus function is most easily understood if you separate the function into its different sections and give them rules.

First consider the function within the modulus. Where is it positive? Where is it negative?

For the positive parts, the original function will hold, i.e. you just need to remove the modulus sign from the function given.

For the negative parts, you need to negate the original function (apply neg sign once modulus is taken away) as you are reflecting it in the x-axis to make those y-values positive.

Then you've got your rules - it's easier to do anything with modulus functions now.