Find the points on the curve y=((x^3)/3)-x where the tan line is parallel to the line y=3x?

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  • david
    Lv 7
    3 weeks ago
    Favourite answer

     y= ((x^3)/3) - x 

    y' = x^2 - 1 = 3

      x^2 - 4 = 0

       x = +2  and   -2

      (2, 2/3)  and  (-2, -2/3)  <<< 2 points

  • 3 weeks ago

    y'(x) = x^2 - 1

    The line y = 3x has a slope of 3.

    x^2 - 1 = 3

    x^2 = 4

    x = -2, 2

    Points on curve where tangent line is parallel to the line y = 3x are (-2, -2/3) and (2, 2/3).

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