Answer
$$\eqalign{
& {\text{max}} = \sqrt 2 {\text{ at }}x = \frac{{3\pi }}{4} \cr
& {\text{min}} = - 1{\text{ at }}x = 0 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sin x - \cos x;\,\,\left[ {0,\pi } \right] \cr
& f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr
& {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr
& {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr
& {\text{the open interval }}\left( {0,\pi } \right){\text{.}} \cr
& \cr
& {\text{Differentiating }}f\left( x \right) \cr
& f'\left( x \right) = \cos x + \sin x \cr
& {\text{setting }}f'\left( x \right) = 0 \cr
& \cos x + \sin x = 0 \cr
& {\text{For the interval }}\left[ {0,\pi } \right]{\text{ we obtain }} \cr
& x = \frac{{3\pi }}{4} \cr
& {\text{Thus}}{\text{, there is a stationary points at }}x = \frac{{3\pi }}{4} \cr
& {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = - \frac{\pi }{4},{\text{ }}x = \frac{\pi }{2}{\text{ and the}} \cr
& {\text{stationary point}}{\text{.}} \cr
& f\left( 0 \right) = \sin \left( 0 \right) - \cos \left( 0 \right) = - 1 \cr
& f\left( {\frac{{3\pi }}{4}} \right) = \sin \left( {\frac{{3\pi }}{4}} \right) - \cos \left( {\frac{{3\pi }}{4}} \right) = \sqrt 2 \cr
& f\left( \pi \right) = \sin \left( \pi \right) - \cos \left( \pi \right) = 1 \cr
& {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr
& {\text{the }}\,{\text{interval }}\,\,\left[ {0,\pi } \right]{\text{ is}}\,{\text{ }} - 1{\text{ at }}x = 0.{\text{ }}\,{\text{And}}\,{\text{ the }}\,{\text{absolute}} \cr
& {\text{maximum is }}\sqrt 2 {\text{ at }}x = \frac{{3\pi }}{4} \cr
& \cr
& {\text{max}} = \sqrt 2 {\text{ at }}x = \frac{{3\pi }}{4} \cr
& {\text{min}} = - 1{\text{ at }}x = 0 \cr} $$