#### Answer

Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-\dfrac{1}{2}.$

#### Work Step by Step

$f'(x)=\cos{x}.$
$f'(x)=0\to x=\dfrac{\pi}{2}.$
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 $-\dfrac{1}{2}.$