Well suppose for an example of an inequality,
|x-1|-|x|+|2x+3| > 2x+4

Well in one of its solutions we were told to apply the method of intervals, rather than taking say what; like 8 combination of signs.

For everyone of its intervals(say -3/2 $\leq$x <0) we are said that 2x+3 $\geq$ 0, x<0 and x-1<0.

Is it just an intuitive outcome(guessing by the extremities of the limits) or did we do something to predict what the signs of the expressions will be?(as i was wondering if any such intervals, which violated an intuitive outcome ?)

HallsofIvy
Homework Helper
I'm not sure what you mean by "intuitive" but the method of "predicting" the sign is to use the definition of "absolute value": |x|= x if $x\ge 0$, -x if x< 0.

Of course that says that the "change" comes when the argument of the absolute value function is 0. In order to deal with |x-1|-|x|+|2x+3| > 2x+4, I would note that x-1= 0 when x=1, x= 0 when x= 0, of course, and 2x+3= 0 when x= -3/2. Now put those in order: -3/2< 0 < 1. So if x< -3/2, it is also less than 0 and 1 and all of x-1, x and 2x+3 are negative. For x< -3/2, the inequality becomes -(x- 1)-(-x)+ (-(2x+3))> 2x+ 4. That is the same as -x+ 1+ x- 2x- 3= -2x- 2> 2x+ 4 which is, in turn, equivalent to -6> 4x or -3/2< x. Of course, if x< -3/2, "-3/2< x" can't be true so there is NO value of x, less than -3/2, satisfying that inequality.

If $-3/2\le x< 0$, x and x- 1 are still negative but 2x+3 is now non-negative and so the inequality is -(x-1)-(-x)+ (2x+3)> 2x+ 4. That is the same as -x+ 1+ x+ 2x+ 3= 2x+ 4> 2x+4 which is, again, never true (they equal not "larger than"). There is no x between -3/2 and 0 satifying this inequality.

If $0\le x< 1$ both 2x+3 and x are positive but x- 1 is still negative. Now the unquality is -(x-1)- x+ (2x+3)> 2x+ 3. That is the same as -x+ 1+ x+ 2x+ 3= 2x+ 4> 2x+ 3 which is always true. Every value of x, $0\le x< 1$ satisfies this inequality.

Finally, if $x\ge 1$ all three of x-1, x, and 2x+3 are positive. Now the inequality is (x- 1)- x+ (2x+ 3)= 2x+ 2> 2x+ 3. That is never true.

Putting all of those together, the original inequality is satified for all x, $0\le x< 1$ and no other x. The solution set is the interval [0, 1).