Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}:y=7x^3+11 \\ \therefore \frac{dy}{dx}=21x^2 \\ \mathrm{Now}, \\ \text{Slope of the tangent at (}x=2)={\left(\frac{dy}{dx}\right)}_{x=2}=21{\left(2\right)}^2=84 \\ \text{Slope of the tangent at (}x=-2)\text{=}{\left(\frac{dy}{dx}\right)}_{x=-2}=21{\left(-2\right)}^2=84 \end{gathered}$$

Both slopes are the same. Hence, the tangents at points x = 2 and x = −2 are parallel.