Thread: solution set of |x-1|+|x+1| < 6

1. solution set of |x-1|+|x+1| < 6

Hi, I am having some problems with an exercise.

I have solved the first one, but I don't know how to go about the 2nd one .

The first one is:
Find the solution set of |x-1|+|x+1|<1

And I got this:
|x|=
-2x, x<=-1
2x, x>=1
2, -1<x<1
(I got this from drawing the graph for |x-1| and |x+1| and adding the parallel lines)

For the second exercise I have:
Find the solution set of
|x-1|+|x+1|<6

I am pretty sure the result cannot be the same from looking at the graph, but I have no clue how to work it out using calculus :-/.

Any help would be very much appreciated, thanks!

2. $|x-1|+|x+1| < 6$

$x-1 =0 \Rightarrow x = 1$

$x+1 =0 \Rightarrow x = -1$

meaning you need to look at the regions $(-\infty , -1 ) \cup (-1, 1) \cup (1 , +\infty)$ of the region $(-\infty , +\infty)$

meaning you say now
1st case is $x\in (-\infty , -1 )$ so you will have $(x-1)<0$ and $(x+1) <=0$ so equation will be

$-(x-1) - (x+1) < 6$

solve than you see if solution is in that interval ? if not than that is not the solution (or solutions) than next interval and so on ... later solutions are all that satisfy conditions

3. no calculus need to solve this.

we just focus on where the absolute values get zero, that happens when $x=1$ and $x=-1,$ note that both points satisfy the inequality, hence, in order to remove the absolute value, we study the cases when $(-\infty,-1],\,[-1,1],[1,\infty).$