Answer
$u = \pm \sqrt 3 $
Work Step by Step
$$\eqalign{
& B\left( u \right) = 4{\tan ^{ - 1}}u - u \cr
& {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr
& {\text{Differentiate with respect to }}u \cr
& B'\left( u \right) = \frac{d}{{du}}\left[ {4{{\tan }^{ - 1}}u - u} \right] \cr
& B'\left( u \right) = 4\left( {\frac{1}{{{u^2} + 1}}} \right) - 1 \cr
& B'\left( u \right) = \frac{4}{{{u^2} + 1}} - 1 \cr
& B'\left( u \right) = \frac{{4 - {u^2} - 1}}{{{u^2} + 1}} \cr
& B'\left( u \right) = \frac{{3 - {u^2}}}{{{u^2} + 1}} \cr
& {\text{The derivative }}B'\left( u \right){\text{ exists for all real number }}u,{\text{ then}} \cr
& {\text{the critical numbers occur when }}B'\left( u \right) = 0 \cr
& \frac{{3 - {u^2}}}{{{u^2} + 1}} = 0 \cr
& 3 - {u^2} = 0 \cr
& - {u^2} = - 3 \cr
& {u^2} = 3 \cr
& u = \pm \sqrt 3 \cr} $$
