1. ## Find the domain

Hi, im trying to find the domain of $f(x)=arcsin(1/x)$

I know that $y=arcsinx$ is defined on the interval (-1,1),
so when I look at f(x) then (1/x) should be on the interval:

$-1<=(1/x)<=1$

I know that the solution is $x>=1$ and $x<=-1$ (x is bigger or equal to 1, and less or equal to -1) But I dont know how they got there? any help? thanks!

2. ## Re: Find the domain

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).

3. ## Re: Find the domain

You know that $0<\frac{1}{x}\le{1}$ if and only if $x \ge 1$, and $-1\le\frac{1}{x}<0$ if and only if $x \le -1$. There is no real number $x$ that makes $\frac{1}{x}$ zero. So your condition on $\frac{1}{x}$ and the answer have the same meaning.

- Hollywood

4. ## Re: Find the domain

As I said the correct answer should be $x>=1$ and $x<=-1$ 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:
$-1<=(1/x)<=1$

Then: (multiplying all sides with x)
$-x<=1<=x$

Then I have:

$-x<=1$ and $1<=x$

which is equal to (?):

$x>=-1$ and $1<=x$ 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

5. ## Re: Find the domain

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?

6. ## Re: Find the domain

Originally Posted by Duffman

If:
$-1<=(1/x)<=1$

Then: (multiplying all sides with x)
$-x<=1<=x$
When you multiply both sides of an inequality by a negative number, the direction of the inequality is reversed. So you have $-x \ge 1$ instead of $-x \le 1$.

- Hollywood

7. ## Re: Find the domain

Im sorry I seem slow

But when I have
$-1 \le (1/x)\le 1$
... and multiply both sides with $x$, isnt $x$ 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?
$-x \ge (1)\le x$

Best regards
/Duffman

8. ## Re: Find the domain

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

9. ## Re: Find the domain

Originally Posted by Duffman
Hi, im trying to find the domain of $f(x)=arcsin(1/x)$

I know that $y=arcsinx$ is defined on the interval (-1,1),
so when I look at f(x) then (1/x) should be on the interval:

$-1<=(1/x)<=1$
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, $(-\infty, -1)\cup (1, \infty)$.

I know that the solution is $x>=1$ and $x<=-1$ (x is bigger or equal to 1, and less or equal to -1) But I dont know how they got there? any help? thanks!