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Thread: find how many solutions this ln polynomial equation has..

  1. #1
    MHF Contributor
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    find how many solutions this ln polynomial equation has..

    $\displaystyle
    2x^3lnx+x^2-4x+a=0
    $
    i thought of building a taylor series around 0 for ln
    but ln(0) is undefined

    ??
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let $\displaystyle f(x)=2x^3\ln x+x^2-4x+a, \ f0,\infty)\to\mathbf{R}$

    We use the Rolle's sequence.

    $\displaystyle f'(x)=6x^2\ln x+2x^2+2x-4$

    $\displaystyle f'(x)=0\Rightarrow x=1$ and f' is increasing so it has an unique solution.

    $\displaystyle \lim_{x\searrow 0}f(x)=a, \ f(1)=a-3, \ \lim_{x\to\infty}f(x)=\infty$

    If $\displaystyle a\in(-\infty,0]$ then the Rolle's sequence is $\displaystyle - \ - \ +$ and f has a solution $\displaystyle x_1\in(1,\infty)$

    If $\displaystyle a\in(0,3)$ then the Rolle's sequence is $\displaystyle + \ - \ +$ and f has two solutions $\displaystyle x_1\in(0,1), \ x_2\in(1,\infty)$

    If $\displaystyle a=3$ then the Rolle's sequence is $\displaystyle + \ 0 \ +$ and f has the solution $\displaystyle x=1$

    If $\displaystyle a>3$ then the Rolle's sequence is $\displaystyle + \ + \ +$ and f has no real solution.
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