Show that the tangents to the curve $y = 2 x^{3} - 3$ at the points where $x = 2$ and $x = - 2$ are parallel.

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Solution

The equation of the curve is $y = 2 x^{3} - 3$ .

Differentiating w.r.t. $x$ , we get,

$\frac{d y}{d x} = 6 x^{2}$

Now, $m_{1} =$ (slope of tangent $x = 2) = {(\frac{d y}{d x})}_{x = 2} = 6 \times (2)^{2} = 24$

and $m_{2} =$ (slope of tangent at $x = - 2) = {(\frac{d y}{d x})}_{x = - 2} = 6 (- 2)^{2} = 24$

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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