# 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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