# The points at which the tangents to the curve $y = x^3 - 12x + 18$ are parallel to x-axis are: A. (2, -2), (-2, -34) B. (2, 34), (-2, 0) C. (0, 34), (-2, 0) D. (2, 2), (-2, 34)

D)

Given the equation of the curve is

$y = x^3 - 12x + 18$

Differentiating on both sides with respect to x, we get

$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}-12 \mathrm{x}+18\right)}{\mathrm{dx}}$
Applying the sum rule of differentiation, we get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-12 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(18)}{\mathrm{dx}}$
We know derivative of a constant is 0,so above equation becomes
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-12 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+0$
Applying the power rule we get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-12 \ldots(i)$
Thus, the slope of line parallel to the x -axis is given by
$\frac{\mathrm{dy}}{\mathrm{dx}}=0$
So equating equation (i) to 0 we get
$3 x^{2}-12=0$

$\\ { \Rightarrow 3x\textsuperscript{2}=12}\\ { \Rightarrow x\textsuperscript{2}=4}\\ { \Rightarrow x= \pm 2}\\$

When x=2, the given equation of curve becomes,

$\\ {y = x\textsuperscript{3} - 12x + 18}\\ { \Rightarrow y = (2)\textsuperscript{3} - 12(2) + 18}\\ { \Rightarrow y = 8- 24 + 18}\\ { \Rightarrow y = 2}\\$

When x=-2, the given equation of curve becomes,

$\\ {y = x\textsuperscript{3} - 12x + 18}\\ { \Rightarrow y = (-2)\textsuperscript{3} - 12(-2) + 18}\\ { \Rightarrow y = -8+24 + 18}\\ { \Rightarrow y = 34}\\$

Hence, the points at which the tangents to the curve $y = x^3 - 12x + 18$ are parallel to x-axis are (2, 2) and (-2, 34).

So, the correct option is option D.