1. ## Inequality assistance

The question:

Solve $\displaystyle |\frac{x - 1}{x + 1}| < 1$

My attempt:

$\displaystyle |x - 1| < |x + 1|$
$\displaystyle x - 1 < x + 1$
$\displaystyle x < x + 2$

Help! I'm clearly doing something wrong here. :/

2. Originally Posted by Glitch
The question:

Solve $\displaystyle |\frac{x - 1}{x + 1}| < 1$

My attempt:

$\displaystyle |x - 1| < |x + 1|$
$\displaystyle x - 1 < x + 1$
This doesn't follow. If x- 1> 0 and x+ 1> 0 (which happens for x> 1), then |x- 1|= x- 1 and |x+ 1|= x+ 1 so x- 1< x+ 1 from which -1< 1 which is always true.

But if x- 1 and x+ 1 are both negative (which happens for x< -1), then |x- 1|= -(x- 1) and |x+ 1|= -(x+ 1) so -(x- 1)< -(x+ 1) from which x- 1> x+ 1, -1> 1 which is always false.

If -1< x< 1, then x- 1< 0 and x+ 1> 0 so |x- 1|< |x+ 1| becomes -(x- 1)< x+ 1 or -x+ 1< x+ 1 which reduces to -x< x, 0< 2x, x> 0.

The inequality is true for all positive x except x= 1. (If the problem had been $\displaystyle \frac{|x- 1|}{|x+ 1|}\le 1$ then it would be true for all positive x.)

$\displaystyle x < x + 2$

Help! I'm clearly doing something wrong here. :/

3. A rule!

If you see |f(x)|<t it's : -t<f(x)<t

And if you see |f(x)|>t then it's:

f(x)>t or f(x)<-t

In your case it's the first one when f(x)=(x-1)/(x+1)

4. Originally Posted by Also sprach Zarathustra
A rule!

If you see |f(x)|<t it's : -t<f(x)<t

And if you see |f(x)|>t then it's:

f(x)>t or f(x)<-t

In your case it's the first one when f(x)=(x-1)/(x+1)
Ahh, thanks. I'll remember that.

I'm still having trouble simplifying it. I've done:

$\displaystyle -1 < |\frac{x - 1}{x + 1}| < 1$
$\displaystyle -(x + 1) < (x - 1) < (x + 1)$
$\displaystyle (-x - 1) < (x - 1) < (x + 1)$
$\displaystyle -x < x < x + 2$
$\displaystyle 0 < 2x < 2x + 2$
$\displaystyle 0 < x < x + 1$

So x > 0 (which is correct).

But am I doing this correctly? Or was it just a coincidence I got the answer?

5. Originally Posted by Glitch
Ahh, thanks. I'll remember that.

I'm still having trouble simplifying it. I've done:

$\displaystyle -1 < |\frac{x - 1}{x + 1}| < 1$
$\displaystyle -(x + 1) < (x - 1) < (x + 1)$
$\displaystyle (-x - 1) < (x - 1) < (x + 1)$
$\displaystyle -x < x < x + 2$
$\displaystyle 0 < 2x < 2x + 2$
$\displaystyle 0 < x < x + 1$

So x > 0 (which is correct).

But am I doing this correctly? Or was it just a coincidence I got the answer?
Incorrect! (first line)

$\displaystyle -1 < \frac{x-1}{x+1} < 1$

If so, does this mean I have to square the denominator to ensure it's positive before solving?

7. Yes!

And how you solve this?

In fact you have two inequalities:

1. (x-1)/(x+1)<1

AND

2. (x-1)/(x+1)>-1

I will start the first one:

(x-1)/(x+1)<1

==> (x-1)/(x+1)-1<0

==> (x-1 -x-1)/(x+1)<0

==> -2/(x+1)<0

And now like you said , square the denominator.

8. But I can't do anything with that square, because if I multiply each side by (x + 1)^2 then the RHS is 0(x + 1)^2, which is 0. Doing so gave me x > -1

Sorry if I'm doing it wrong, I'm having a bad day I think...

9. Ok. I will continue from

==> -2/(x+1)<0

==> -2(x+1)/(x+1)^2<0

==> -2(x+1)<0 (because (x+1)>0 fo all x in R\(-1) )

==> -(x+1)<0

==> x+1>0

==> x>-1

Now try the second one.

10. Ok, the second one gives me x > 0. So since, x > -1 and x > 0, I naturally take x > 0 (since 0 > -1), right?

11. Yes!

12. Thank you good sir. I've just started my calculus course, and these pre-calc questions are worrying me!

13. You welcome! And good-luck!