Step 1: Function Analysis.
Given \( f(x) = \sin x + \cos x \) on \( [0, \pi] \), we rewrite \( f(x) \).Using trigonometric identity:\[f(x) = \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right)\]Step 2: Determine Extreme Values.
Since the range of \( \sin \left( x + \frac{\pi}{4} \right) \) is \( [-1, 1] \):\[\text{Maximum of } f(x) = \sqrt{2} \times 1 = \sqrt{2}\]\[\text{Minimum of } f(x) = \sqrt{2} \times (-1) = -\sqrt{2}\]Step 3: Adjust for Interval.
On \( [0, \pi] \), \( \sin \left( x + \frac{\pi}{4} \right) \) achieves its maximum at \( x = \frac{\pi}{4} \). The minimum would occur at \( x = \frac{5\pi}{4} \), but this is outside the interval. The minimum value within \( [0, \pi] \) is \( -1 \) because the lowest value of \( \sin(x+\frac{\pi}{4}) \) in the interval is \( \sin(\pi) = \sin(\frac{5\pi}{4}) = -1 \).Therefore, the maximum is \( \sqrt{2} \) and the minimum is \( -1 \). Final Answer: \[ \boxed{\sqrt{2} \text{ and } -1} \]