Answer
Relative minimum: $\left(\frac{3 \pi}{4},-1\right)$
Relative maximum: $\left(\frac{\pi}{4}, 1\right)$
Work Step by Step
First derivative
\[
2 \cos 2 x=f^{\prime}(x)
\]
Critical points (zeros of first derivative or point where first derivative does not exist)
\[
x=\frac{3 \pi}{4} \text { and } x=\frac{ \pi}{4}
\]
Using the first derivative test to determine relative extrema.
$f^{\prime}(x)<0$ on the left and $f^{\prime}(x)>0$ on the right of $x=\frac{3 \pi}{4},$ so $x=\frac{3 \pi}{4}$ is a relative minimum.
$f\left(\frac{3 \pi}{4}\right)=-1$
$f^{\prime}(x)>0$ on the left and $f^{\prime}(x)<0$ on the right of $x=\frac{\pi}{4},$ so $x=\frac{\pi}{4}$ is a relative maximum.
$f\left(\frac{\pi}{4}\right)=1$
Second derivative
\[
-4 \sin 2 x=f^{\prime \prime}(x)
\]
Using the second derivative test to determine relative extrema.
$f^{\prime}\left(\frac{3 \pi}{4}\right)=0$ and $f^{\prime \prime}\left(\frac{3 \pi}{4}\right)=4>0,$ so $x=\frac{3 \pi}{4}$ is a relative minimum.
$f^{\prime}\left(\frac{\pi}{4}\right)=0$ and $f^{\prime \prime}\left(\frac{\pi}{4}\right)=-4<0,$ so $x=\frac{\pi}{4}$ is a relative maximum.
