calculus

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find the absolute minimum value on (0,infinity)...f(x)= x-(1/x)+(2/x^3)..

The example given is x-(1/x)+(10/x^3) which is F(sqrt5)=(6/sqrt5)

  • calculus -

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

    f'=0 when x^4 + 2x^2 - 6 = 0
    x^2 = -1 ± √7

    -1 - √7 < 0. so we have

    x^2 = -1 + √7
    x = ±√(√7 - 1)

    we want x in (0,∞), so

    f(1.28287) = 1.451

    or, more exactly,

    f(√(√7 - 1)) = √((97√7 - 143)/54)

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