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Thread: Find the values of x

  1. #1
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    Find the values of x

    such that $\displaystyle -2 < x^3 - 2x^2 + x - 2 < 0 $.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    $\displaystyle -2<x^3-2x^2+x-2<0$

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

    $\displaystyle x^2+1$ is positive

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

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

    $\displaystyle 0<x<2$
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by mark1950 View Post
    such that $\displaystyle -2 < x^3 - 2x^2 + x - 2 < 0 $.
    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.

    Quote Originally Posted by alexmahone View Post
    $\displaystyle -2<(x-2)(x^2+1)<0$

    $\displaystyle x^2+1$ is positive

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

    $\displaystyle -2<(x-2)<0$
    $\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$
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  4. #4
    MHF Contributor alexmahone's Avatar
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by TheAbstractionist View Post


    $\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$
    I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.
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  6. #6
    Senior Member TheAbstractionist's Avatar
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    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.
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  7. #7
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    Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mark1950 View Post
    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.
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  9. #9
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    Quote Originally Posted by alexmahone View Post
    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:
    $\displaystyle 0<2x<1$
    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.
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  10. #10
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by pfarnall View Post
    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.
    Fair enough. Sorry.
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