Find the local maximum and local minimum values for the function $f(x) = \sin x - \cos x$ where $0 < x < 2\pi$.

Given $f(x) = \sin x - \cos x$ for $0 < x < 2\pi$.
First,find the derivative $f'(x) = \cos x + \sin x$.
Set $f'(x) = 0$ to find critical points: $\cos x + \sin x = 0 \Rightarrow \tan x = -1$.
In the interval $(0, 2\pi)$,$\tan x = -1$ at $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.
Now,find the second derivative $f''(x) = -\sin x + \cos x$.
Evaluate at $x = \frac{3\pi}{4}$: $f''(\frac{3\pi}{4}) = -\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0$. Thus,$x = \frac{3\pi}{4}$ is a point of local maxima.
The local maximum value is $f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) - \cos(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \sqrt{2}$.
Evaluate at $x = \frac{7\pi}{4}$: $f''(\frac{7\pi}{4}) = -\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0$. Thus,$x = \frac{7\pi}{4}$ is a point of local minima.
The local minimum value is $f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) - \cos(\frac{7\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$.