Question:
Who that the tangents to the curve $y=7 x^{3}+11$ at the points $x=2$ and $x=-2$ are parallel.
Solution:
Given: $y=7 x^{3}+11$
$\therefore \frac{d y}{d x}=21 x^{2}$
Now,
Slope of the tangent at $(x=2)=\left(\frac{d y}{d x}\right)_{x=2}=21(2)^{2}=84$
Slope of the tangent at $(x=-2)=\left(\frac{d y}{d x}\right)_{x=-2}=21(-2)^{2}=84$
Both slopes are the same. Hence, the tangents at points $x=2$ and $x=-2$ are parallel.
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