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October 31, 2014

October 31, 2014

Posted by **Susana** on Monday, October 3, 2011 at 12:54pm.

- Calculus -
**Reiny**, Monday, October 3, 2011 at 2:46pmFor a horizontal tangent dy/dx = 0

dy/dx = √3 - 2sinx = 0

2sinx = √3

sinx = √3/2

x = 60° or 120°

x = π/3 or 2π/3

when x=π/3

y = √3(π/3) + 2cos(π/3) = √3(π/3) + 2(1/2) = (π√3 + 3)/3

when x = 2π/3

y = √3(2π/3) + 2(-1/2) = 2√3 π - 3)/3

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