Show that the tangents to the curve `y=7x^3+11` at the points `x=2` and `x=-2` are parallel.

Explanation

The equation of the given curve is `y=7 x^{3}+11`.

After differentiating

`\therefore \frac{d y}{d x}=21 x^{2}`


Thus slope of the tangent to the curve at `x=2` is `=(\frac{d y}{d x})_{x=2}=(21 x^{2})_{x=2}=84`


and slope of the tangent to the curve at `x=-2` is `=(\frac{d y}{d x})_{x=-2}=(21 x^{2})_{x=-2}=84`


As we can see that slope of both the tangents are equal.

Hence, the two tangents are parallel.