help modulus equations/graphs

-|x+1|+|1-x|=2

what calculations you need to do to find the answer?

If x<-1 will either/neither/both of your modulus be negative? and if so ... what would you have to do

What about if x>1

What about if -1<x<1



Do you now why I used these values?

so if I use x=-1, the answer will be 2, if i use x&gt;1, the answer will be negative; and if I use -1&lt;x&lt;1, the answers could be negative or positive... so I don't know what would be the answer

I'll give you an example.

Say you're taking the case x>1. Then x+1>0 and 1-x<0. Therefore |x+1|=x+1 and |1-x|=-(1-x).

So the question simplifies to -(x+1) - (1-x) = 2, therefore -x-1-1+x=2 therefore 2=-2.

This tells you that there are no solutions for x in the range x>1.

Now try the two other cases. If you get any solutions out, make sure thhat they fall in the range for that case and if so, write them down as a possible answer.

You have to try out all three ranges that x could fall in, and see if you can get an answer out.

whenever i get a question like this, I should use the same ranges of x??? x&lt;= -1 ; -1&lt;x&lt;1 ; x&gt;1

You don't use the exact same ranges for every question you get - the reason I chose to have the boundaries of those ranges at x=1 and x=-1 is because the things inside the modulus signs change sign at x=-1 or x=1.

another question:
how would you draw the graph y= Sin|2x|

When $x\geq 0$ then |2x|=2x so sin|2x| = sin(2x).

When x<0 then |2x| = -2x so sin|2x|=sin(-2x).

Therefore, you need to take the graph of y=sin(2x) and make it so that the part for x<0 is the same as the part for x>0 but reflected in the y axis.

the change for the modulus would be the red bit ???

Almost - the loop at the left end of the curve should be pointing downwards. It would look a bit like this:

should'nt it be a stretch of 1/x in the x-axis???