|x+1|+|x-1|<=2
this means that sum of distance of x from -1 and from 1 is less or equal to 2.
|x+1| distance will be either x+1 or -x-1
|x-1| distance will be either x-1 or -x+1

so x+1+x-1<=2 => x<=1
and -x-1-x+1<=2 => -2x<=2 => x>=-1

between -1 and 1

rrobiinn wrote:Which values of x are solutions to the inequality [x + 1] + [x - 1] <= 2 ( the sign means less than equal, third bracket means absolute sign.)?
(Hint: try a conceptual approach. Your answer will be an inequality as welL)

Aha, this is also my weakness. But let me give it a shot:

The equation is: |X+1| + |X-1| <=2

Now, the first step would be to determine the critical points.
What do I mean by critical points? The critical points are the values of X that would lead the value of the expression equal to 0. It'd be easier if we apply this

See, the critical points are:

X = -1 and X = 1

How did I arrive to those values? Simple, there X + 1 = 0 as such X = -1 and x - 1 = 0 as such X = 1

Now that we have the critical points, we could go then try values (this is where the conceptual approach comes in handy)

1st step: x < -1 (so we could test x = -2 since -2 < -1)
As such we have:
|-2 + 1| + |-2-1| <=? 2 (mind the "?" as we are determining if the equation holds)
|-1| + |-3| <=? 2
1 + 3 <=? 2
4 > 2 (See, it means that for all values of x <-1, the equation doesn't hold.)

Now on to the next step:

2nd step: -1 <= x <= 1 (this is our critical points)
As such we have:
We try x = 0 (since -1<0<+1)
|0+1| + |0-1| <=? 2 (Again mind the "?")
1 + 1 <=? 2
2 = 2 (equation holds, now let's try another value of x to convince us)
We try x = 1/2
|1/2 + 1| + |1/2 - 1| <=? 2
|3/2| + |-1/2| <=? 2
3/2 + 1/2 <=2

So we see that such equation holds when -1 <= x <= +1

Now the last step would be test x > 2
So we could test x = 3
Let's see,
|3 + 1| + |3 - 1| <=? 2
4 + 2 <=? 2
6 > 2 (doesn't hold.)

Again, we conclude that such equation holds if -1 <= x <= +1