For what values of k equation |x+1|-|x-1|=kx+1 have unique solution?
Since absolute value has two separate formulas depending on whether the number inside the absolute value is positive or negative. That means you need to look at x+1< 0 (x< -1), x+1> 0 (x> -1), x-1< 0 (x< 1), and x-1> 0 (x> 1).
That means we need to look at three intervals: x< -1, -1< x< 1, and x> 1.
If x< -1, both x+1< 0 and x-1< 0 so the equation becomes -(x+1)-(-(x-1))= -2= 2k+1 which has no solution for x if x is not -3/2 but an infinite number of solutions if k= 3/2 (since then the equation reduces to -2= -2 which is true for all x).
If -1< x< 1, x+1> 0 but x-1< 0 so the equation becomes (x+1)-(-(x-1))= 2x= 2k+1 which has a single solution for x no matter what k is.
If x> 1, both x+1>0 and x-1> 0 so the equation becomes (x+1)- (x-1)= 2= 2k+1 which has no solution for x is k is not 1/2 but has an infinite number of solutions if k= 1/2 (because then the equation reduces to 2= 2 which is true for all x).
You said that there is infinite number of solutions if k=3/2 but i find only one solution and that is -2. abs(x+1)-abs(x-1)=(3/2)x+1 - Wolfram|Alpha
My fault, I misread the equation! I accidently replaced the "x" with "2".
For x< -1, the left side, as I said, reduces to -2 so the equation becomes -2= kx+ 1. Now subtract 1 from each side to get kx=-3 and divide by x to get x= -3/k, as long as k is not 0. If k= 0, this gives no solution.
For -1< x< 1, the left side reduces to 2x so the equation becomes 2x= kx+ 1. subrtract kx from both sides to get 2x-kx= (2- k)x= 1 and then x= 1/(2- k) as long as k is not 2. If k= 2, this gives no solution.
For x> 1, the left side reduces to 2 so the equation becomes 2=kx+ 1. subtract 1 from both sides to get kx= 1 and divide by x to get x= 1/k, as long as k is not 0. If k= 0, ths gives no solution.
Now, what must k be so that only one of those gives a solution?