Answer :

Given,

Now put f^{’}(x) = 0,

⇒ 3x^{6} – 3 = 0

⇒ x^{6} – 1 = 0

⇒ x^{6} = 1

Therefore, above equation has only two solutions

x = 1 and x = -1

For x < -1 and x > 1

Put x = 2, we have

And for x ∈ (-1, 1)\{0}

Put x = 1/2

Now, As f’(x) < 0 in (-1, 1)\{0}, f(x) is strictly decreasing in (-1, 1)\{0}

And as f(x) > 0 in (-∞, -1) ∪ (1, ∞), f(x) is strictly increasing in (-∞, -1) ∪ (1, ∞)

**OR**

We know, derivative of a function at any point gives the slope of tangent at that point.

Now, given tangent

y = x – 11

comparing with, y = mx + c

we get, m = 1 (slope)

Now,

For, m = 1

⇒ 1 = 3x^{2} – 11

⇒ 3x^{2}= 12

⇒ x^{2} = 4

⇒ x = ± 2

When x = 2

⇒ y = x – 11 = 2 – 11 = -9

When x = -2

⇒ y = x – 11 = -2 – 11 = -13

Therefore, possible points are (2, -9) and (-2, -13)

Now, equation of curve y = x^{3} – 11x + 5

At x = 2

y = (2)^{3} – 11(2) + 5 = -9

∴ (2, -9) satisfy the curve equation

At x = -2

y = (-2)^{3} – 11(-2) + 5

⇒ y = -8 + 22 + 5 = 19

⇒ (-2, -13) doesn’t satisfy the equation

∴ (2, -9) is the required point!

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