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

1.	Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.
Answer» Slope of the line y = x – 11 is 1 slope of the curve \(\frac{dy}{dx}\)= 3x² – 11 ∴ 3x² – 11 = 1 3x² = 12 ⇒ x² = 4, x = +2 when x = 2, y = (2)³ – 11 (2) + 5 = -9 when x = -2, y = -8 + 22 + 5 = 19 points are (2, -9) and (-2, 19) equation of tangent at (2, -9) and slope is 1 y + 9 = 1 (x – 2) y = x- 11 equation of tangent at (-2, 19) y – 19 = 1 (x + 2) ⇒ y = x + 211 ∴ (2, -9) is the only point at which the tangent is y = x – 11.10.

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

Answer»

Slope of the line y = x – 11 is 1

slope of the curve \(\frac{dy}{dx}\)= 3x² – 11

∴ 3x² – 11 = 1

3x² = 12 ⇒ x² = 4, x = +2

when x = 2, y = (2)³ – 11 (2) + 5 = -9

when x = -2, y = -8 + 22 + 5 = 19

points are (2, -9) and (-2, 19)

equation of tangent at (2, -9) and slope is 1

y + 9 = 1 (x – 2)

y = x- 11 equation of tangent at (-2, 19)

y – 19 = 1 (x + 2)

⇒ y = x + 211

∴ (2, -9) is the only point at which the tangent is y = x – 11.10.