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Find the point on the curve y = x^3-11x + 5 at which the tange

Question

Find the point on the curve $y = x^{3} - 11 x + 5$ at which the tangent is $y = x - 11.$

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Evelyn Discussed

Find the point on the curve

y = x^{3} - 11 x + 5

at which the tangent is

y = x - 11.

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

The equation of the given curve is

y = x^{3} - 11 x + 5

The equation of the tangent to the given curve is given as

y = x - 11

(which is of the form

y = m x + c

∴

Slope of the tangent

= 1

Now, the slope of the tangent to the given curve at the point

(x, y)

is given by,

\frac{d y}{d x} = 3 x^{2} - 11

Then, we have:

3 x^{2} - 11 = 1 \Rightarrow 3 x^{2} = 12

\Rightarrow x^{2} = 4

\Rightarrow x = \pm 2

When

x = 2, y = (2)^{3} - 11 (2) + 5 = 8 - 22 + 5 = - 9

When

x = - 2, y = (- 2)^{3} - 11 (- 2) + 5 = - 8 + 22 + 5 = 19

Hence, the required points are

(2, 9)

and

(2, 19) .

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Question Text	Find the point on the curve $y = x^{3} - 11 x + 5$ at which the tangent is $y = x - 11.$
Answer Type	Text solution:1
Upvotes	150