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TANGENTS AND NORMALS
find a point on the curve y=x^3 – 3x where the tangent is parallel to the chord joining (1, -2) and (2, 2)

taniska , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$y = x^3 - 3x$
$y' = 3x^2 - 3$
Slope of tangent:
$m = \frac{2-(-2)}{2-1}$
$m = 4$
$3x^2 - 3 = 4$
$3x^2 = 7$
$x = \pm \sqrt{\frac{7}{3}}$
Put in the curve
$y = \pm ((\sqrt{\frac{7}{3}})^{3}-(\sqrt{\frac{7}{3}}))$
$y = \pm \sqrt{\frac{7}{3}}(\frac{7}{3}-1)$
$y = \pm \frac{4}{3}\sqrt{\frac{7}{3}}$

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