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Thread: Inflection point

  1. #1
    VkL
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    Inflection point

    f(x) = (1+x) / (1 + x^2)

    what are the inflection points?

    I got f '(x) = - (x^2 + 2x -1) / (x^4 + 2x + 1)
    I got f ''(x) = (2x^3 + 6x^2 - 6x - 2) / (x^6 + 3x^4 + 3x^2 + 1)

    Now When I set f ''(x) = zero to find the inflection point(s), How can I solve for x....this is...amazingly hard.
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    Quote Originally Posted by VkL View Post
    f(x) = (1+x) / (1 + x^2)

    what are the inflection points?

    I got f '(x) = - (x^2 + 2x -1) / (x^4 + 2x + 1)
    I got f ''(x) = (2x^3 + 6x^2 - 6x - 2) / (x^6 + 3x^4 + 3x^2 + 1)

    Now When I set f ''(x) = zero to find the inflection point(s), How can I solve for x....this is...amazingly hard.
    Remember that a rational function can never have 0 in the denominator. To find the 0 of f''(x), all you need to do is set the numerator equal to zero. Once you,ve done this, find the roots of the cubic polynomial.
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  3. #3
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    Quote Originally Posted by VkL View Post
    f(x) = (1+x) / (1 + x^2)

    what are the inflection points?

    I got f '(x) = - (x^2 + 2x -1) / (x^4 + 2x + 1)
    I got f ''(x) = (2x^3 + 6x^2 - 6x - 2) / (x^6 + 3x^4 + 3x^2 + 1)

    Now When I set f ''(x) = zero to find the inflection point(s), How can I solve for x....this is...amazingly hard.
    f''(x) = \frac{2(x^3+3x^2-3x-1)}{(x^2+1)^3}<br />

    note the numerator ...

    2(x^3 + 3x^2 - 3x - 1)

    using the rational root theorem, x = 1 turns out to be a zero, resulting in the following factorization ...

    \frac{2(x - 1)(x^2+4x+1)}{(x^2+1)^3} = 0

    x = 1 , x = -2 \pm \sqrt{3}
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