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calculus

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Determine the coordinates of the point on the graph of f(x)= (2x+1)^(1/2) where the tangent line is perpendicular to the line 3x+y+4=0

  • calculus -

    dy/dx = (1/2) (2x+1)^(-1/2) (2)
    = 1/(2x+1)^(1/2) or 1/√(2x+1)

    slope of line is -3
    so slope of perpendicular is 1/3

    so we have ...
    1/√(2x+1) = 1/3
    √(2x+1) = 3
    square both sides ...
    2x + 1 = 9
    2x = 8
    x = 4

    f(4) = (9)^1/2) = 3
    so the point of contact is (4,3)
    and the equation of the tangent is

    y-3 = (1/3)(x-4)
    3y - 9 = x-4
    x - 3y = -5 OR y = (1/3)x + 5/3

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