The absolute maximum and minimum values of the function $f(x) = \sin x + \sqrt{3}\cos x$ in $[0, \pi]$ are

Given $f(x) = \sin x + \sqrt{3}\cos x$

We can rewrite the function as:

$f(x) = 2 \left( \dfrac{1}{2} \sin x + \dfrac{\sqrt{3}}{2} \cos x \right) = 2 \sin\left(x + \dfrac{\pi}{3}\right)$

For $x \in [0, \pi]$, the angle $\left(x + \dfrac{\pi}{3}\right) \in \left[\dfrac{\pi}{3}, \dfrac{4\pi}{3}\right]$.

The maximum value of $\sin\left(x + \dfrac{\pi}{3}\right)$ in this interval is $1$, which occurs at $x + \dfrac{\pi}{3} = \dfrac{\pi}{2} \Rightarrow x = \dfrac{\pi}{6}$.

Thus, the absolute maximum value of $f(x)$ is $2 \times 1 = 2$.

The minimum value of $\sin\left(x + \dfrac{\pi}{3}\right)$ in this interval occurs at the endpoint $x + \dfrac{\pi}{3} = \dfrac{4\pi}{3} \Rightarrow x = \pi$.

Thus, the absolute minimum value of $f(x)$ is $2 \sin\left(\dfrac{4\pi}{3}\right) = 2 \left(-\dfrac{\sqrt{3}}{2}\right) = -\sqrt{3}$.

Alternatively, evaluating the function at the critical point $x = \dfrac{\pi}{6}$ and the endpoints $x = 0, \pi$:

$f(0) = \sqrt{3}$

$f\left(\dfrac{\pi}{6}\right) = 2$

$f(\pi) = -\sqrt{3}$

Therefore, the absolute minimum value is $-\sqrt{3}$ and the absolute maximum value is $2$.

Answer: Minimum value $= -\sqrt{3}$, maximum value $= 2$