Question

find the point on the curve at which the tangent to the curve y=3x^2-2x+1 is parallel to x Axis​

Answer 1

Answer:

Let (x,y) be the required point.

Let (x,y) be the required point.The slop of line y=3x+4 is 3

Let (x,y) be the required point.The slop of line y=3x+4 is 3Since, the point lie

2

2 −x+1

2 −x+1Differentiate w.r.t x,

2 −x+1Differentiate w.r.t x,⇒

2 −x+1Differentiate w.r.t x,⇒ dx

2 −x+1Differentiate w.r.t x,⇒ dxdy

2 −x+1Differentiate w.r.t x,⇒ dxdy

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent =

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dx

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And⇒ y=2x

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And⇒ y=2x 2

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And⇒ y=2x 2 −x+1=2(1)

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And⇒ y=2x 2 −x+1=2(1) 2

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And⇒ y=2x 2 −x+1=2(1) 2 −1+1=2

2 −x+1Differentiate w.r.t x,⇒ dxdy =4x−1Now,Slope of the tangent = dxdy =4x−1Slop of the tangent = Slop of the given lien [ Given ]∴ 4x−1=3⇒ 4x=4⇒ x=1And⇒ y=2x 2 −x+1=2(1) 2 −1+1=2∴ Required point is (1,2)