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November 28, 2014

November 28, 2014

Posted by **Tina** on Friday, November 26, 2010 at 5:51pm.

f(θ) = 2cos(θ) + (sin(θ))^2

- Calculus -
**Reiny**, Friday, November 26, 2010 at 10:49pmlet's take f '(Ø)

f '(Ø) = -2sinØ + 2sinØcosØ

= 0 for max/mins

sinØ - sinØcosØ = 0

sinØ(1 + cosØ) = 0

sinØ = 0 or cosØ = -1

Ø = 0,π,2π or Ø = 3π/2

f(0) = 2+0 = 2

f(π) = -2+0 = -2

f(2π) = 2 + 0 = 2

f(3π/2) = 0 + 1 = 0

max value of the function is 2 when Ø = 0 or 2π

min value of the function is -2 when Ø = 3π/2

- Calculus -
**Anonymous**, Friday, October 26, 2012 at 9:44am2sinθcosθ-1=0

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