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Determine the Relative Extrema of the Function y=2cosx+sin2xy = 2 \cos x + \sin 2 x

Question 48

Multiple Choice

Determine the relative extrema of the function y=2cosx+sin2xy = 2 \cos x + \sin 2 x on the interval (0,2π) ( 0,2 \pi ) .


A) relative minimum: (5π6,332) \left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right) relative maximum: (5π6,332) \left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)
B) relative minimum: (5π6,332) \left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right) relative maximum: (π6,332) \left( \frac { \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)
C) relative minimum: (π6,332) \left( \frac { \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right) relative maximum: (5π6,332) \left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)
D) relative minimum: (π6,332) \left( \frac { \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right) relative maximum: (5π6,332) \left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)
E) relative minimum: (5π6,332) \left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right) relative maximum: (π6,332) \left( \frac { \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)

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