Find the point on the curve y = x 3 – 11x + 5 at which the tangent is y = x –11.

Question

Philoid · Accepted Answer

It is given that equation of the curve y = x³ – 11x + 5

At which the tangent is y = x –11

⇒ Slope of the tangent = 1

Now, slope of the tangent to the given curve at a point (x,y) is:

⇒ 3x² -11 = 1

⇒ 3x² = 12

⇒ x² = 4

⇒ x = 2

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

And when x = -2 then y = (-2)³ -11(-2) + 5 = 19

Therefore, required points are (2, -9) and (-2, 19).