The points at which the tangents to the curve $y = x^{3} - 12 x + 18$ are parallel to x-axis are?

(2, - 2), (- 2, - 34)

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(2, 34), (- 2, 0)

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(0, 34), (- 2, 0)

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(2, 2), (- 2, 34)

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Solution

The correct option is D $(2, 2), (- 2, 34)$
Given curve $y = x^{3} - 12 x + 18$
on differentiating the above equation of the curve we get
$\frac{d y}{d x} = 3 x^{2} - 12$
If the tangent is parallel to x axis $\frac{d y}{d x} = 0$
$\Rightarrow 3 x^{2} - 12 = 0$
$3 x^{2} = 12$
$x^{2} = 12 / 3$
$x^{2} = 4$
$x = \pm 2$
Now x = 2, $y = (2)^{3} - 12 (2) + 18$
$= 8 - 24 + 18$
$= 26 - 24 = 2$
When x = -2 $y = (- 2)^{3} - 12 (- 2) + 18$
$= - 8 + 24 + 18$
$42 - 8 = 34$
Hence at (2,2) & (-2,34) the tangent to the curve is parallel to x axis

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