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

(N/A) The equation of the given curve is $y=7x^3+11$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 21x^2$.
The slope of the tangent to a curve at a point $(x_0, y_0)$ is given by $\left. \frac{dy}{dx} \right|_{(x_0, y_0)}$.
For $x=2$,the slope of the tangent is:
$m_1 = \left. \frac{dy}{dx} \right|_{x=2} = 21(2)^2 = 21 \times 4 = 84$.
For $x=-2$,the slope of the tangent is:
$m_2 = \left. \frac{dy}{dx} \right|_{x=-2} = 21(-2)^2 = 21 \times 4 = 84$.
Since the slopes of the tangents at $x=2$ and $x=-2$ are equal $(m_1 = m_2 = 84)$,the tangents are parallel.