Determine the relative extrema of the function $y = 2 \cos x + \sin 2 x$ on the interval $( 0,2 \pi )$ .
A)relative minimum: $\left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)$ relative maximum: $\left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)$
B)relative minimum: $\left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)$ relative maximum: $\left( \frac { \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)$
C)relative minimum: $\left( \frac { \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)$ relative maximum: $\left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)$
D)relative minimum: $\left( \frac { \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)$ relative maximum: $\left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)$
E)relative minimum: $\left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)$ relative maximum: $\left( \frac { \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)$

Question

Quizplus · Accepted Answer

To find the relative extrema, we first need to find the derivative of the function $y = 2 \cos x + \sin 2x$, which is $y' = -2 \sin x + 2 \cos 2x$. Setting $y'$ equal to zero gives the critical points. Simplifying and solving for $x$ within the given interval $(0, 2\pi)$ will lead to finding the correct critical points. Evaluating the second derivative or using a sign chart for the first derivative around these critical points will help determine whether they are relative maxima or minima. The correct critical points and their classification as minima or maxima correspond to choice B.

Determine the Relative Extrema of the Function y=2cos⁡x+sin⁡2xy = 2 \cos x + \sin 2 xy=2cosx+sin2x

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