The points at which the tangents to the curve $$y=x^3-12x+18$$ are parallel to x-axis are?
Given curve $$y=x^{3}-12x+18$$
on differentiating the above equation of the curve we get
$$\frac{dy}{dx}=3x^{2}-12$$
If the tangent is parallel to x axis $$\frac{dy}{dx}=0$$
$$\Rightarrow 3x^{2}-12=0$$
$$3x^{2}=12$$
$$x^{2}=12/3$$
$$x^{2}=4$$
$$x=\pm 2$$
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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