For the function $f(x) = sinx + cosx, x \in [0, \pi]$, which one of the following is correct?

For the function $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$, the absolute maximum and minimum occur either at critical points (where the derivative equals zero) or at the endpoints of the interval.

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Taking the derivative:

$f'(x) = \cos x - \sin x$

Setting $f'(x) = 0$:

$\cos x - \sin x = 0$

$\cos x = \sin x$

$\tan x = 1$

$x = \frac{\pi}{4}$

The critical point is $x = \frac{\pi}{4}$.

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Evaluating at $x = 0$:

$f(0) = \sin 0 + \cos 0$

$f(0) = 0 + 1$

$f(0) = 1$

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Evaluating at $x = \frac{\pi}{4}$:

$f\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4}$

$f\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$

$f\left(\frac{\pi}{4}\right) = \frac{2}{\sqrt{2}}$

$f\left(\frac{\pi}{4}\right) = \sqrt{2}$

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Evaluating at $x = \pi$:

$f(\pi) = \sin \pi + \cos \pi$

$f(\pi) = 0 + (-1)$

$f(\pi) = -1$

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Comparing the values:

At $x = 0$: $f(0) = 1$

At $x = \frac{\pi}{4}$: $f\left(\frac{\pi}{4}\right) = \sqrt{2}$

At $x = \pi$: $f(\pi) = -1$

The absolute maximum value is $\sqrt{2}$ occurring at $x = \frac{\pi}{4}$.

The absolute minimum value is $-1$ occurring at $x = \pi$.

Therefore, the correct option is: The absolute maximum value is $\sqrt{2}$.