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Math Help - Solutions of functions

  1. #1
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    Solutions of functions

    How would you approach these questions:

    1)\ Show\ that\ x^3+x-9=0\ has\ only\ one\ real\ solution.

    Would you differentiate it and then show that the derivative must be greater than 0 for all values of x. Then find a point, which gives a negative value of f(x) and a point that gives a positive value of f(x).

    2)\ How\ many\ real\ zeros\ does\ p(x)=x^3-12x^2+45x-51\ have?

    Would the answer always be 3. Could there be a case, where it is not 3.
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  2. #2
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    q1

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    Last edited by Neverquit; April 24th 2010 at 10:04 AM. Reason: x
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  3. #3
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    q2

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    Last edited by Neverquit; April 24th 2010 at 10:05 AM. Reason: x
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  4. #4
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    q2

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    Last edited by Neverquit; April 24th 2010 at 10:05 AM. Reason: x
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    q1

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    Last edited by Neverquit; April 24th 2010 at 10:05 AM. Reason: x
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  6. #6
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    Quote Originally Posted by acevipa View Post
    How would you approach these questions:

    1)\ Show\ that\ x^3+x-9=0\ has\ only\ one\ real\ solution.

    Would you differentiate it and then show that the derivative must be greater than 0 for all values of x. Then find a point, which gives a negative value of f(x) and a point that gives a positive value of f(x).

    2)\ How\ many\ real\ zeros\ does\ p(x)=x^3-12x^2+45x-51\ have?

    Would the answer always be 3. Could there be a case, where it is not 3.
    For 1, I would do as you said - the derivative is 3x^2 + 1 > 0 \ \forall x \in \mathbb{R}. Now all you need to do is find one point where f is negative and one where it is positive.

    For 2, an example of a 3rd degree polynomial which doesn't have 3 real roots is (x+i)(x-i)(x+1) = (x^2 + 1)(x + 1) = x^3 + x^2 + x + 1, which has one real root = -1 and two complex roots -  \pm i

    In this example, differentiate the polynomial and find how many zeroes the derivative has (this is easier - it's just a quadratic in x..), and then how many extremal points. This should tell you how many zeroes the polynomial has (why?)
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