# 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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- davidLv 73 weeks agoFavourite 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

- stanschimLv 73 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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