math

If f(x)=4x-(1/x) when x>0 and g is the inverse of f, then what is the value of g'(0)?

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asked by Anonymous

  1. f(x)=4x-(1/x)
    or
    y = 4x - 1/x

    step 1 of inverse: interchange the x and y's
    x = 4y - 1/y
    multiply by y
    xy = 4y^2 - 1 <----- the g relation
    when x = 0, 0 = 4y^2 - 1 ---> y = 1/2
    xy' + y = 8y y'
    xy' - 8y y' = -y
    y'(x-8y) = -y
    y' = -y/(x-8y) = y/(8y - x) <----- g'
    plug in x=0, y=1/2
    g'(0) = (1/2) / (8(1/2))
    = 1/8

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    posted by Reiny

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