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

Solution:

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The given curve is

y = x³ - 11x + 5

The equation of the tangent to the given curve is y = x - 11

Therefore,

the slope of the tangent is 1.

Now, the slope of the tangent to the given curve is,

dy/dx

= d/dx (x³ - 11x + 5)

= 3x² - 11

Hence,

3x² - 11 = 1

⇒ 3x² = 12

⇒ x² = 4

⇒ x = ± 2

When, x = 2

Then,

y = (2)³ - 11(2) + 5

= 8 - 22 + 5

= - 9

When, x = - 2

Then,

y = (-2)³ - 11(- 2) + 5

= - 8 + 22 + 5

= 19

Thus, the points are (2, - 9) and (- 2, 19)

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 9

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

Summary:

The point on the curve y = x³ - 11x + 5 at which the tangent is y = x - 11 are (2, - 9) and (- 2, 19). The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line