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