Answer
$$x = \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& y = \sqrt x \left( {x - 1} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt x \left( {x - 1} \right)} \right] \cr
& {\text{Using the product rule}} \cr
& \frac{{dy}}{{dx}} = \sqrt x \frac{d}{{dx}}\left[ {x - 1} \right] + \left( {x - 1} \right)\frac{d}{{dx}}\left[ {\sqrt x } \right] \cr
& \frac{{dy}}{{dx}} = \sqrt x + \left( {x - 1} \right)\left( {\frac{1}{{2\sqrt x }}} \right) \cr
& {\text{The tangent line to the graph is horizontal when }}\frac{{dy}}{{dx}} = 0,{\text{ then}} \cr
& \sqrt x + \left( {x - 1} \right)\left( {\frac{1}{{2\sqrt x }}} \right) = 0 \cr
& {\text{Solve for }}x \cr
& \sqrt x + \frac{x}{{2\sqrt x }} - \frac{1}{{2\sqrt x }} = 0 \cr
& \sqrt x + \frac{1}{2}\sqrt x - \frac{1}{{2\sqrt x }} = 0 \cr
& \frac{3}{2}\sqrt x - \frac{1}{{2\sqrt x }} = 0 \cr
& \frac{3}{2}\sqrt x = \frac{1}{{2\sqrt x }} \cr
& \frac{2}{3}\sqrt x \left( {\frac{3}{2}\sqrt x } \right) = \frac{2}{3}\sqrt x \left( {\frac{1}{{2\sqrt x }}} \right) \cr
& x = \frac{1}{3} \cr} $$
