QPaperGen

Let
\[
f(x) = \sin x + \cos x
\]

Differentiate:
\[
f'(x) = \cos x - \sin x
\]

Set \( f'(x) = 0 \):
\[
\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1
\Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4}, \ldots
\]

Now compute the second derivative:
\[
f''(x) = -\sin x - \cos x = -(\sin x + \cos x)
\]

To determine if it's a maximum, evaluate at \( x = \frac{\pi}{4} \):
\[
f''\left( \frac{\pi}{4} \right) = -\left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) = -\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) = -\sqrt{2} < 0
\]

Since \( f''(x) < 0 \), this is a local maximum.

Now compute the value:
\[
f\left( \frac{\pi}{4} \right) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}
\]

\[
\boxed{\text{Maximum value of } \sin x + \cos x \text{ is } \sqrt{2}, \text{ at } x = \frac{\pi}{4}}
\]