# Find the point on

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).

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