Absolute value inequality

  • Thread starter cragar
  • Start date
  • #1
2,544
2

Homework Statement


|1/(x-1)|<1

The Attempt at a Solution


is that the same as this -1<1/(x-1)<1

can i do each side by it self then take the values at which they intersect

so i subtracted the 1 then got (-x+2)/(x-1) then made a sign chart with 2 and 1 on it
then took the less than terms so i got (1,2) for the first suoltions
then for the second i got x/(x-1) then made a sign chart with 0 and 1 on it
then i took the greater than terms geting (-inf,0) U (1,inf)
is this right
 

Answers and Replies

  • #2
56
0
is that the same as this -1<1/(x-1)<1
Well, let's check. Suppose x = 0. Does this inequality hold? No because -1 = -1.

Maybe this will help:

|x| < y
=> x < y or -x > -y
 
  • #3
2,544
2
ok i see so you took x<y then divided it by a -1 and then flipped the sign
so then would our soultions to my original inequality be
(-inf,1) U (2,inf)
 
  • #4
statdad
Homework Helper
1,495
35
A shorter idea (for future reference)
[tex]
|x| < y
[/tex]

is the same as
[tex]
-y < x < y
[/tex]

If your problem begins with [tex] \le [/tex] then replace [tex] < [/tex] with [tex] \le [/tex] in the simplification.
 
  • #5
2,544
2
would we then have to do a sign chart
 
Last edited:
  • #6
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,987
1,085
crager,
Look again at Dunkle's first response. Graph the relation on a number line and you may more clearly find a path to a solution of his example and to your exercise problem.

In your original expression on the left, the expression inside of the absolute value is either positive, or negative; examine each of these conditions separately. Do you yet need more detailed descriptions?
 
  • #7
2,544
2
(-inf,-2) U (2,inf) is this the answer
 
  • #8
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,987
1,085
Reread post #6 and #2. Check your answer.
 
  • #9
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,987
1,085
Start with |y|<1, where y = 1/(x-1).

Use a number line to help understand this:
Either y<1 OR -y<1.

Replace y with the original expression and solve, and check or first solve for y and then replace y with the original expression and finish solving.
...but be careful. I'm not absolutely sure about my own answer, so I might have made a conceptual error. Still, not that for the particular exercise, x canNOT be 1.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
956
In my opinion, the simplest way to solve most inequalities is to first solve the associated equation. |1/(x-1)|= 1 reduces to 1/(x-1)= 1 or 1/(x-1)= -1. Multiplying both sides of each gives 1= x-1 and 1= 1- x. In the first case, x= 2 and in the second x= 0. The point of that is that continuous functions can change from "<" to ">" on where they are "=". Of course, |1/(x-1)| is NOT continuous at x= 1 so we must add that possibility: the inequality can change at x= 0, x= 1, and x= 2. If we take x= -1< 0, |1/(-1-1)|= 1/2< 1 so the given inequality is true for all x< 0. Check a value of x between 0 and 1, a value of x between 1 and 2, and a value of x larger than 2 to determine which of those intervals also satisfy the inequality.
 

Related Threads on Absolute value inequality

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
559
  • Last Post
Replies
5
Views
869
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
15K
Top