Question

Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.

Answer 1

Slope of the line y = x – 11 is 1.

Given; 

y = x³ – 11x + 5 ⇒ \(\frac{dy}{dx}\) = 3x² – 11 = 1

⇒ 3x² = 12 ⇒ x = ±2

At x = 2, ⇒ y = x – 11 = 2- 11 = -9

⇒ (2, -9)

At x = -2, ⇒ y = x – 11 = -2 – 11 = -13

⇒ (-2, -13)

But the point (-2, -13) do not lie on the curve, hence the point is (2, -9).