Show that the tangents to the curve y = 7x3 + 11 at the points x = 2 a

1.	Show that the tangents to the curve y = 7x3 + 11 at the points x = 2 and x = – 2 are parallel.
Answer» Given: The curve y = 7x³ + 11 Differentiating the above w.r.t x ⇒ \(\frac{dy}{dx}\) = 3 x 7x ^{3 – 1} + 0 ⇒ \(\frac{dy}{dx}\) = 21x² when x = 2 ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21×(2)² ⇒ \(\cfrac{dy}{dx_{x=2}}\)= 21x 4 ⇒ \(\cfrac{dy}{dx_{x=2}}\)= 84 when x = – 2 ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x (-2)² ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x 4 ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 84 Let y = f(x) be a continuous function and P(x₀,y₀) be point on the curve, then, The Slope of the tangent at P(x,y) is f'(x) or \(\frac{dy}{dx}\) 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.

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

Answer»

Given:

The curve y = 7x³ + 11

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\) = 3 x 7x ^{3 – 1} + 0

⇒ \(\frac{dy}{dx}\) = 21x²

when x = 2

⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21×(2)²

⇒ \(\cfrac{dy}{dx_{x=2}}\)= 21x 4

⇒ \(\cfrac{dy}{dx_{x=2}}\)= 84

when x = – 2

⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x (-2)²

⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x 4

⇒ \(\cfrac{dy}{dx_{x=2}}\) = 84

Let y = f(x) be a continuous function and P(x₀,y₀) be point on the curve, then,

The Slope of the tangent at P(x,y) is f'(x) or \(\frac{dy}{dx}\)

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.