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For this question why do i have to look at the regions of the modulus function?
And how do i do it?
Do i consider each modulus graph and find the region in which it acts?

Thanks!

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Reply 1
Original post by Zenarthra


For this question why do i have to look at the regions of the modulus function?
And how do i do it?
Do i consider each modulus graph and find the region in which it acts?

Thanks!


Well, how else would you do it? :biggrin:

The point is that if you look at different regions you can replace modulus signs by + or - as appropriate e.g. if you have something like |x + 1| then in the region x < - 1 this is -(x + 1) whereas elsewhere it is just x + 1.
Original post by davros
Well, how else would you do it?


prsom
Reply 3
Original post by davros
Well, how else would you do it? :biggrin:

The point is that if you look at different regions you can replace modulus signs by + or - as appropriate e.g. if you have something like |x + 1| then in the region x < - 1 this is -(x + 1) whereas elsewhere it is just x + 1.




Ahh ok, but i dont see why the modulus function is determined under these domains?

Thanks!
Reply 4
Original post by Zenarthra


Ahh ok, but i dont see why the modulus function is determined under these domains?

Thanks!


You seem to have misunderstood. You can't look at the terms separately.

I'll get you started. How can we rewrite the equation if we know that x>2?
Reply 5
Original post by BabyMaths
You seem to have misunderstood. You can't look at the terms separately.

I'll get you started. How can we rewrite the equation if we know that x>2?


Im guessing to combine it all onto 1 side but im not sure why. :frown:
Am i right? :redface:
Reply 6
Original post by Zenarthra
Im guessing to combine it all onto 1 side but im not sure why. :frown:
Am i right? :redface:


The point is that when x2x\ge 2 we have

|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2

so the equation can be written as..?
Reply 7
Original post by BabyMaths
The point is that when x2x\ge 2 we have

|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2

so the equation can be written as..?


Im sorry but how did you deduce that if x> or = 2 then
|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2 ?

Thanks!
Reply 8
If x>2 then x+1>3 and so |x+1|=x+1.
Original post by Zenarthra
Im sorry but how did you deduce that if x> or = 2 then
|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2 ?

Thanks!

When x>2, all those expressions have a value greater than or equal to 0, so are not reflectes in the x-axis. Therefore the modulus graph of y=x-2 when x>2 has equation y=x-2 and not y=-(x-2) or*2-x. This is true for all the expressions, so once you've stated your range considerations, you can effectively ignore the mod signs and solve. :smile:
Reply 10
And I thought this was going to lead into a Beyonce song. Shame.
Reply 11
Original post by PhysicsKid
When x>2, all those expressions have a value greater than or equal to 0, so are not reflectes in the x-axis. Therefore the modulus graph of y=x-2 when x>2 has equation y=x-2 and not y=-(x-2) or*2-x. This is true for all the expressions, so once you've stated your range considerations, you can effectively ignore the mod signs and solve. :smile:


Ahh right i understand now, but how would i go about solving the equation?

Thanks!
Reply 12
Original post by BabyMaths
If x>2 then x+1>3 and so |x+1|=x+1.


Ahh ok, but i still dont know how i could re write the equation, could you show me please so i can figure out what has happened?

Would it be (x+1)-x+3(x-1)-2(x-2)-(x+2)=0 ?
(edited 9 years ago)
Original post by Zenarthra
Ahh right i understand now, but how would i go about solving the equation?

Thanks!


The mod signs serve as brackets, so expand, collect like terms and it's then just a case of solving a linear equation :smile:
You have to consider the regions in which the different signs act on

When x > 2, no mod signs are in effect

What about between 2 and 1? 1 and 0?

Posted from TSR Mobile
Reply 15
Original post by PhysicsKid
The mod signs serve as brackets, so expand, collect like terms and it's then just a case of solving a linear equation :smile:


(x+1)-x+3(x-1)-2(x-2)-(x+2)=0
If i expand out and simplify dont i get x-6 = x+2?
Reply 16
Original post by Arithmeticae
You have to consider the regions in which the different signs act on

When x > 2, no mod signs are in effect

What about between 2 and 1? 1 and 0?

Posted from TSR Mobile


In those further region there are mods signs that act, but how does this help me solve it?
And also what is the effect of adding mod equations, ofr instance what would |x+1|-|x| look like?
Just a straight line but reflected at the part where it hits the x axis?

Thanks!
Reply 17
Nvm i got it now, ty.
Reply 18
Original post by Zenarthra
In those further region there are mods signs that act, but how does this help me solve it?
And also what is the effect of adding mod equations, ofr instance what would |x+1|-|x| look like?
Just a straight line but reflected at the part where it hits the x axis?

Thanks!


I would suggest forgetting about graphs and just follow the algebra through!

When x > 0 then |x| = x and |x + 1| = x + 1 so |x + 1| - |x| = x + 1 - x = 1

Now deal with the other regions similarly on a case by case basis.
Reply 19
Original post by davros
I would suggest forgetting about graphs and just follow the algebra through!

When x > 0 then |x| = x and |x + 1| = x + 1 so |x + 1| - |x| = x + 1 - x = 1

Now deal with the other regions similarly on a case by case basis.


Yeah i understand now, just out of curiosity what would this graph look like?

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