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Math Help - A Simple Question

  1. #1
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    A Simple Question

    which I cannot do :P

    Well, I just dont remember the best method to answer it

    " f(x)=x^3+8x-19

    show that the equation f(x)=0 has only one real root."

    Thanks a lot
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by Silver View Post
    which I cannot do :P

    Well, I just dont remember the best method to answer it

    " f(x)=x^3+8x-19

    show that the equation f(x)=0 has only one real root."

    Thanks a lot
    f'(x) = 3x^2 + 8 > 0

    Thus f(x) is an increasing function. Now by intermediate value theorem \exists c \in (1,2): f(c) = 0.

    Since the function is increasing, \forall x > c, f(x) > 0; \forall x < c, f(x) < 0 .

    Thus f(x) \neq 0, x \neq c and hence f(x) has only one real root.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Silver View Post
    which I cannot do :P

    Well, I just dont remember the best method to answer it

    " f(x)=x^3+8x-19

    show that the equation f(x)=0 has only one real root."

    Thanks a lot
    By Descartes rule of signs it has at most 1 positive real root.

    Also changing the sign of all odd power terms and applying Descartes rule of signs agrain we see that it has at most 0 negative real roots.

    But this is a cubic and so has at leat one real root, and so it must have exactly one real root and its positive.

    RonL
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