Dayaan M. answered • 7d
Earned A’s Twice in Precalculus | 5 Years of Tutoring Experience
The trick with a nested absolute value like this is to stop trying to handle it all at once and instead break it at the point where the inside absolute value changes its mind, which is at x = 1. On each side of that the inside piece becomes something ordinary, and the problem gets much smaller.
Start with x ≥ 1. There |x - 1| is just x - 1, so the inside of the outer bars becomes x - (x - 1), which is simply 1. The x terms cancel completely, so the whole left side is |1| = 1, no matter which x we picked. So on this side the equation is asking:
floor(x) = 1
and floor(x) equals 1 exactly when 1 ≤ x < 2. That whole interval works, which is a little surprising the first time you see it.
Now take x < 1. There |x - 1| flips to 1 - x, so the inside becomes x - (1 - x) = 2x - 1, and the left side is |2x - 1|. Here is where the floor function does us a favor. An absolute value can never be negative, so floor(x) cannot be negative either, which forces x ≥ 0. Combined with x < 1, that traps us in 0 ≤ x < 1, and on that interval floor(x) is 0. So we need:
|2x - 1| = 0
which happens only at x = 1/2, and that value is inside the interval we were allowed, so it counts.
It is worth testing a couple of these. At x = 1.5 we get |1.5 - 0.5| = 1 and floor(1.5) = 1, so that checks out. At x = 1/2 we get |0.5 - 0.5| = 0 and floor(0.5) = 0. And x = 2 fails, since the left side is still 1 but floor(2) = 2, which is why the interval stops just short of 2.
So, in interval notation, our solution is {1/2} ∪ [1, 2).
