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:
