Q. 94.4( 42 Votes )
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x –11.
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).
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