Show that the tangents to the curve y = \(2x^3 – 3\) at the points where x =2 and x = -2 are parallel.

Solution :

The equation of the curve is y = \(2x^3 – 3\)

Differentiating with respect to x, we get

\(dy\over dx\) = \(6x^2\)

Now, \(m_1\) = (Slope of the tangent at x = 2) = \(({dy\over dx})_{x = 2}\) = \(6 \times (2)^2\) = 24

and, \(m_2\) = (Slope of the tangent at x = -2) = \(({dy\over dx})_{x = -2}\) = \(6 \times (-2)^2\) = 24

Clearly \(m_1\) = \(m_2\).

Thus, the tangents to the given curve at the points where x = 2 and x = -2 are parallel.


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