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calculus

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find an equation of the tangent line to the curve y = x √x that is parallel to the line y = 1 + 3x.

  • calculus -

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

    slope of line = 3

    so x/(2√x) + √x = 3
    x/(2√x) = 3 - √x
    square both sides
    x^2/(4x) = 9 - 6√x + x
    times 4
    x/4 = 9 - 6√x + x
    simplifying ...
    8√x = 12+x
    square again
    64x = 144 + 24x + x^2
    x^2 - 40x + 144 = 0
    (x-36)(x-4) = 0
    x = 36 or x = 4

    there will be two tangents possible
    I will do one, you do the other.

    if x=36, y = 216
    let y = 3x + b
    216 = 3(36) + b
    b = 108

    y = 3x + 108

    etc.

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