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

Answers (1)

Slope of tangent = \frac{dy}{dx} = 21x^2 
When x = 2
\frac{dy}{dx} = 21x^2 = 21(2)^{2} = 21 \times4 = 84
When  x = -2
\frac{dy}{dx} = 21x^2 = 21(-2)^{2} = 21 \times4 = 84
Slope is equal when x= 2 and x = - 2 
Hence, we can say that both the tangents to curve y = 7x^3 + 11 is parallel

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