Thread: Inequality assistance

The question:

Solve

My attempt:





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

Originally Posted by Glitch

The question:

Solve

My attempt:

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 then it would be true for all positive x.)

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

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)

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:








So x > 0 (which is correct).

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

Originally Posted by Glitch

Ahh, thanks. I'll remember that.

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








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)

Please read again 'my rule'...

So it's this instead?



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

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.

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...

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.

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

Yes!

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

You welcome! And good-luck!