Find the point on the curve y = x3 where the Slope of the tangent is equal to x – coordinate of the point.

Given:


The curve is y = x3


y = x3


Differentiating the above w.r.t x


= 3x2 – 1


= 3x2 ...(1)


Also given the The Slope of the tangent is equal to the x – coordinate,


= x ...(2)


From (1) & (2),we get,


i.e, 3x2 = x


x(3x – 1) = 0


x = 0 or x =


Substituting x = 0 or x = this in y = x3,we get,


when x = 0


y = 03


y = 0


when x =


y = )3


y =


Thus, the required point is (0,0) & (,)


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