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

The equation of the given curve is y = x³ − 11x + 5.

The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c).

∴ Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point (x, y) is given by, = 6(x + 1)² (x – 3)² (x – 1)

Then, we have:

f’ (x) = 0 ⇒ x = -1, 3, 1

When x = 2, y = (2)³ − 11 (2) + 5 = 8 − 22 + 5 = −9.

When x = −2, y = (−2)³ − 11 (−2) + 5 = −8 + 22 + 5 = 19.

Hence, the required points are (2, −9) and (−2, 19).

But point (- 2, 19) does not satisfy the equation of tangent. 

Therefore, required point is (2, - 9).