Show that the tangents to the curve

Solution:

Given:

The curve $y=7 x^{3}+11$

Differentiating the above w.r.t $\mathrm{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \times 7 \times^{3-1}+0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=21 \times^{2}$

when $x=2$'

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}_{\mathrm{x}}=2}=21 \times(2)^{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}_{\mathrm{x}}=2}=21 \times 4$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}_{\mathrm{x}}=2}=84$

when $x=-2$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}_{\mathrm{x}}=2}=21 \times(-2)^{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}_{\mathrm{x}}=2}=21 \times 4$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}_{\mathrm{x}}=2}=84$

Let $y=f(x)$ be a continuous function and $P\left(x_{0}, y_{0}\right)$ be point on the curve, then,

The Slope of the tangent at $P(x, y)$ is $f^{\prime}(x)$ or $\frac{d y}{d x}$

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.