## Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-\dfrac{1}{2}.$
$f'(x)=\cos{x}.$ $f'(x)=0\to x=\dfrac{\pi}{2}.$ The interval's endpoints and $x=\dfrac{\pi}{2}$ are possible points at which the function can attain absolute extrema. $f(\frac{5}{6}\pi)=\dfrac{1}{2}.$ $f(\frac{1}{2}\pi)=1.$ $f(\frac{11}{6}\pi)=-\dfrac{1}{2}.$ Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-\dfrac{1}{2}.$