Let $g(x) = f(x) + f(1 - x)$ and $f''(x) > 0, \; x \in (0, 1)$. If $g$ is decreasing in the interval $(0, \alpha)$ and increasing in the interval $(\alpha, 1)$, then $\tan^{-1}(2\alpha) + \tan^{-1}\!\left(\dfrac{1}{\alpha}\right) + \tan^{-1}\!\left(\dfrac{\alpha + 1}{\alpha}\right)$ is equal to: