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Thread: Real analysis - proving the number of solutions

  1. #1
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    Real analysis - proving the number of solutions

    Hey everyone,

    Basically this kinda question comes up every year on the Real Analysis exam, but i really don't know how to do it.
    I'm stuck on the last part of the question with
    f(x) = 1-x.

    PLEASE if anyone can explain to me how you go about doing it...let me know.

    thank you
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  2. #2
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    Quote Originally Posted by kevin28 View Post
    Hey everyone,

    Basically this kinda question comes up every year on the Real Analysis exam, but i really don't know how to do it.
    I'm stuck on the last part of the question with
    f(x) = 1-x.

    PLEASE if anyone can explain to me how you go about doing it...let me know.

    thank you
    For the last part you are looking for the solutions to:

    $\displaystyle x^3-6x^2+12x-8=1-x$

    or:

    $\displaystyle x^3-6x^2+13x-9=0$

    Decartes rule of signs tells us this has $\displaystyle 1$ or $\displaystyle 3$ positive real roots. As the function changes sign between $\displaystyle x=2$ and $\displaystyle x=3$ there is at least one root in this interval.

    Now:

    $\displaystyle \frac{d}{dx}x^3-6x^2+13x-9=3x^2-12x+13$

    Examining the discriminant of the quadratic we find this has no real zeros, and as this is positive for large $\displaystyle x$, the cubic is everywhere an increasing function, and so can have at most one zero, which we have already located between $\displaystyle 2$ and $\displaystyle 3$.


    RonL
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  3. #3
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    thank u

    thank u so much, that was really helpful!
    cheers
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