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


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).


9
1