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) & (,
)