Question

Find the points on the curve y = x³ – 2x² – x where the tangents are parllel to 3x – y + 1 = 0.

Answer 1

Let the required point on the curve
y = x³ – 2x² – x be P(x₁, y₁).
Differentiating y = x³ – 2x² – x w.r.t. x, we get

`dy/dx = d/dx(x^3 - 2x^2 - x)`

= 3x² – 2x 2x – 1
= 3x² – 4x – 1
∴ slope of the tangent at (x₁, y₁)

= `(dy/dx)_("at"(x_1, y_1)`

= 3x₁² – 4x₁ –  1
Since this tangent is parallelto 3x – y + 1 = 0

where slope is `(-3)/(-1)` = 3,

slope of the tangent = 3
∴  3x₁² – 4x₁ – 1 = 3
∴ 3x₁² – 4x₁ – 4 = 0
∴ 3x₁² – 6x₁ + 2x₁ – 4 = 0
∴ 3x₁ (x₁ – 2) + 2(x₁ – 2) = 0
∴ (x₁ – 2)(3x₁ + 2) = 0
∴ x₁ – 2  0 or 3x₁ + 2 = 0
∴ x₁ = 2  x₁ = `-(2)/(3)`
Since, (x₁, y₁) lies on y = x³ – 2x² – x,y₁ = x₁³ – 2x₁² – x₁ 

When x₁ = 2, y₁ = (2)3 – 2(2)2 – 2 = 8 – 8 – 2 = – 2

When x₁ = `-(2)/(3),y_1 = ((-2)/3)^3 - 2((-2)/3)^2 + (2)/(3)`

= `(-8)/(27) - (8)/(9) + (2)/(3)`

= `(-14)/(27)`

Hence the  rquired points are (2, –2) nd `((-2)/3, (-14)/27)`.