First slide

Tangents and normals

Question

A point on the curve $y = x^{3} - 3 x + 5$ at which the tangent line is parallel to $y = - 2 x$ is

Moderate

A

(1, 3)
B

(0, 5)
C

$(\frac{1}{\sqrt{3}}, 5 - \frac{8 \sqrt{3}}{9})$
D

$(\frac{1}{\sqrt{2}}, 0)$

Solution

$Since y^{'} (x) = 3 x^{2} - 3$ so by the condition of parallelism
$3 x^{2} - 3 = - 2$
$x = \pm \frac{1}{\sqrt{3}}$
If $x = \frac{1}{\sqrt{3}}, then y = \frac{1}{3 \sqrt{3}} - \frac{3}{\sqrt{3}} + 5$
$= 5 - \frac{8}{3 \sqrt{3}} = 5 - \frac{8 \sqrt{3}}{9}$

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