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

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- Jul 1st 2009, 05:36 AMmark1950Find the values of x
such that $\displaystyle -2 < x^3 - 2x^2 + x - 2 < 0 $.

- Jul 1st 2009, 05:51 AMalexmahone
$\displaystyle -2<x^3-2x^2+x-2<0$

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

$\displaystyle x^2+1$ is positive

Dividing throughout by $\displaystyle x^2+1$,

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

$\displaystyle 0<x<2$ - Jul 1st 2009, 05:59 AMTheAbstractionist
Concentrate on one side of the inequality at a time.

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

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

Solive this, then go on to the next inequality.

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

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

$\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$ (Shake) - Jul 1st 2009, 06:06 AMalexmahone
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- Jul 1st 2009, 06:09 AMalexmahone
- Jul 1st 2009, 06:11 AMTheAbstractionist
I mean if you divide

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

by $\displaystyle x^2+1$ you should get

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

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

In any case, $\displaystyle x=1$ is not a solution. - Jul 1st 2009, 06:21 AMmark1950
Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?

- Jul 1st 2009, 06:30 AMalexmahone
- Jul 1st 2009, 06:31 AMpfarnall
Think about this example and you'll see you're in error:

$\displaystyle 0<2x<1$

We can all agree that 2 is positive so by your working:

$\displaystyle 0<x<1$

Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term. - Jul 1st 2009, 06:33 AMalexmahone