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Math Help - Show that f(x) + f'(x) + f"(x) > 0 for all real x(Complex numbers)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Show that f(x) + f'(x) + f"(x) > 0 for all real x(Complex numbers)?

    If m^2 + n^2 - mn - m - n - 1\leq 0, where m,n\in\mathbb{R} and z_{1},z_{2} be the complex numbers such that that |z_{1}| \leq m, |z_{2}|\leq n and a > (|z_{1}| - |z_{2}|)^2 + (\arg z_{1} - \arg z_{2})^2 - |z_{1} - z_{2}|^2. If f(x) = ax^2 + bx + c > 0 for all x, show that f(x) + f'(x) + f''(x) > 0 for all real x.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If \color{blue}m^2 + n^2 - mn - m - n - 1\leq 0, where \color{blue}m,n\in\mathbb{R} and \color{blue}z_{1},z_{2} be the complex numbers such that that \color{blue}|z_{1}| \leq m, |z_{2}|\leq n and \color{blue}a > (|z_{1}| - |z_{2}|)^2 + (\arg z_{1} - \arg z_{2})^2 - |z_{1} - z_{2}|^2. If f(x) = ax^2 + bx + c > 0 for all x, show that f(x) + f'(x) + f''(x) > 0 for all real x.
    You have mixed up two separate problems. The sentence in blue is completely irrelevant to the problem contained in the second sentence.

    The conditions for the quadratic polynomial f(x) = ax^2 + bx + c to be always positive are that a>0 and b^2<4ac. If f(x) satisfies those conditions then it easy to check that f(x) + f'(x) + f''(x) = ax^2 + (2a+b)x + (2a+b+c) satisfies the same conditions.
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