Hey Duffman.
Hint: When is |1/x| <= 1 (where |.| is the absolute value)? [Remember that sin(x) always lies in-between -1 and 1 inclusive).
Hi, im trying to find the domain of
I know that is defined on the interval (-1,1),
so when I look at f(x) then (1/x) should be on the interval:
I know that the solution is and (x is bigger or equal to 1, and less or equal to -1) But I dont know how they got there? any help? thanks!
As I said the correct answer should be and which can also be written as (-infinite,-1]u[1,infinite).
Im sorry but I dont fully understand You. This is how I try to solve it:
If:
Then: (multiplying all sides with x)
Then I have:
and
which is equal to (?):
and which can be written as [-1, infinite)]u[1,infinite). (This solution is not OK because it should be (-infinite,-1]u[1,infinite).
I know that something is wrong here but I dont seem to understand what :/ Thanks for our help
Rule: |x|<a if -a<x<a
---------------------
If I try to continue:
-1<=(1/x)<=1
is equal to:
|1/x| <= 1
should be equal to:
1/|x| <= 1
multiplying with |x| on both sides, gives:
1<=|x|
This expression has "two solutions",
x>=1
-x>=1
Can be written as:
x>=1
x<=-1
Which seems to give the correct answer:
(-infinite,-1]u[1,infinite).
My question: Is this method mathematically correct? And can I solve this in any other way?
Im sorry I seem slow
But when I have
... and multiply both sides with , isnt a positive number? So why is the direction of the inequality reversed on the left side? (All I have done is multiplying with positive x).
How do you make it?
Best regards
/Duffman
The problem is that x can represent a negative number. So when you multiply by x you either reverse or don't reverse the direction of the inequality depending on whether x is positive or negative. In other words, there are two cases: x < 0 and x > 0.
- Hollywood
The problem with this is that 0 is between -1 and 1 and 1/x is never 0. Better to say -1<= 1/x< 0 and 0< 1/x<= 1. From -1<= 1/x< 0 we have 1<= -1/x, x<= -1. From 0< 1/x<= 1, 1<= x.
That gives two separate intervals, .
I know that the solution is and (x is bigger or equal to 1, and less or equal to -1) But I dont know how they got there? any help? thanks!