Find the values of x

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July 1st 2009, 05:36 AM
mark1950

Find the values of x

such that $-2 < x^3 - 2x^2 + x - 2 < 0$ .
July 1st 2009, 05:51 AM
alexmahone

$-2<x^3-2x^2+x-2<0$

$-2<(x-2)(x^2+1)<0$

$x^2+1$ is positive

Dividing throughout by $x^2+1$ ,

$-2<(x-2)<0$

$0<x<2$
July 1st 2009, 05:59 AM
TheAbstractionist

Quote:

Originally Posted by mark1950

such that $-2 < x^3 - 2x^2 + x - 2 < 0$ .

Concentrate on one side of the inequality at a time.

$-2 < x^3-2x^2+x-2$

$\implies\ 0<x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$

Solive this, then go on to the next inequality.

$x^3-2x^2+x-2<0$

Solve this (hint: $x-2$ is a factor of the LHS). Then combine the two solutions.

Quote:

Originally Posted by alexmahone

$-2<(x-2)(x^2+1)<0$

$x^2+1$ is positive

Dividing throughout by $x^2+1$ ,

$-2<(x-2)<0$

$-2$ divided by $x^2+1$ is not $-2.$ (Shake)
July 1st 2009, 06:06 AM
alexmahone

-deleted-
July 1st 2009, 06:09 AM
alexmahone

Quote:

Originally Posted by TheAbstractionist

$-2$ divided by $x^2+1$ is not $-2.$ (Shake)

I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.
July 1st 2009, 06:11 AM
TheAbstractionist

I mean if you divide

$-2<(x-2)(x^2+1)<0$

by $x^2+1$ you should get

$\frac{-2}{x^2+1}<x-2<0$

not $-2<x-2<0.$

In any case, $x=1$ is not a solution.
July 1st 2009, 06:21 AM
mark1950

Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?
July 1st 2009, 06:30 AM
alexmahone

Quote:

Originally Posted by mark1950

Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?

That's because x=1 is NOT a solution.
July 1st 2009, 06:31 AM
pfarnall

Quote:

Originally Posted by alexmahone

I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.

Think about this example and you'll see you're in error:
$0<2x<1$
We can all agree that 2 is positive so by your working:
$0<x<1$
Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term.
July 1st 2009, 06:33 AM
alexmahone

Quote:

Originally Posted by pfarnall

Think about this example and you'll see you're in error:
$0<2x<1$
We can all agree that 2 is positive so by your working:
$0<x<1$
Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term.

Fair enough. Sorry.