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If the line 3x-4y = 0 is tangent in the first quadrant to the curve y = x^3 + k, then k is

  • math - ,

    4 y = 3 x + 0
    y = (3/4) x
    slope = m = 3/4

    y = x^3 + k
    dy/dx = slope = 3 x^2
    so
    3 x^2 = 3/4
    x^2 = 1/4
    x = 1/2 (-1/2 is not in first quadrant)
    then
    y = (1/2)^3 + k
    y = 1/8 + k and y = 3x/4
    so
    3 x/4 = 1/8 + k
    but x = 1/2
    3/8 = 1/8 + k
    k = 2/8 = 1/4

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