Question

The points at which the tangents to the curve $y=x^3–12x+18$ are parallel to $x$−axis are

• A. $(0,34)$, $(-2,0)$
• B. $(2,34)$, $(-2,-0)$
• C. $(2,2)$, $(-2,34)$
• D. $(2,-2)$, $(-2,-34)$

Answer 1

The correct option is C $(2,2)$, $(-2,34)$

The given equation of curve is :
$y=x^3-12x+18$
Differentiating the given equation w.r.t. $x$,
$\Rightarrow \frac{dy}{dx}=3x^2-12$

If tangents to the curve are parallel to the $x$ axis, then

$\frac{dy}{dx}=0$

$\Rightarrow 3x^2-12=0$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

When $x=2$,
$y=x^3-12x+18$
$\Rightarrow y=2^3-12\times 2+18=2$

When $x=-2$,
$y=x^3-12x+18$
$=(-2{)}^3-12\times (-2)+18=34$

So, the points are $(2,2)$ and $(-2,34)$
Hence, option $d$ is correct.