Show that the tangents to the curve y = 7x³ + 11 at the points x = 2 and x = – 2 are parallel.

Given:

The curve y = 7x³ + 11

Differentiating the above w.r.t x

⇒ = 37x^{3 – 1} + 0

⇒ = 21x²

when x = 2

⇒ = 21×(2)²

⇒ = 21×4

⇒ = 84

when x = – 2

⇒ = 21×( – 2)²

⇒ = 21×4

⇒ = 84

Let y = f(x) be a continuous function and P(x_0,y₀) be point on the curve, then,

The Slope of the tangent at P(x_,y) is f'(x) or

Since, the Slope of the tangent is at x = 2 and x = – 2 are equal, the tangents at x = 2 and x = – 2 are parallel.