Find the point on the curve y = x3 - 11x+5 at which the equation of ta

1.	Find the point on the curve y = x3 - 11x+5 at which the equation of tangent is y = x - 11.
Answer» Slope of y = x – 11 is equal to 1 m : dy/dx = 3x² - 11 3x² – 11 = 1 ⇒ x = 2 and -2 At x = 2 From the equation of curve, y = (2)³ – 11(2) + 5 = -9 From the equation of tangent, y = 2 – 11 = -9 At x = -2 From the equation of curve, y = (-2)³ – 11(-2) + 5 = 19 From the equation of tangent, y = -2 – 11 = -13 So, the final answer is (2, -9) because at x = -2, y is come different from the equation of curve and tangent which is not possible.

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

Answer»

Slope of y = x – 11 is equal to 1

m : dy/dx = 3x² - 11

3x² – 11 = 1

⇒ x = 2 and -2

At x = 2

From the equation of curve, y = (2)³ – 11(2) + 5 = -9

From the equation of tangent, y = 2 – 11 = -9

At x = -2

From the equation of curve, y = (-2)³ – 11(-2) + 5 = 19

From the equation of tangent, y = -2 – 11 = -13

So, the final answer is (2, -9) because at x = -2, y is come different from the equation of curve and tangent which is not possible.