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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)

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) 

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 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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