Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 43

$$\eqalign{ & {\bf{a}}.\,\left( {4,4} \right) \cr & {\bf{b}}.\left( {16,0} \right) \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = 4\sqrt x - x \cr & {\bf{a}}.\, \cr & {\text{Differentiate }}f\left( x \right) \cr & {\text{ }}f'\left( x \right) = 4\left( {\frac{1}{{2\sqrt x }}} \right) - 1 \cr & {\text{ }}f'\left( x \right) = \frac{2}{{\sqrt x }} - 1 \cr & {\text{Set }}f'\left( x \right) = 0{\text{ and solve for }}x \cr & \,\frac{2}{{\sqrt x }} - 1 = 0 \cr & \,\frac{2}{{\sqrt x }} = 1 \cr & \sqrt x = 2 \cr & \,\,\,\,x = 4 \cr & {\text{Evaluate }}f\left( 4 \right) \cr & f\left( 4 \right) = 4\sqrt 4 - 4 \cr & f\left( 4 \right) = 4 \cr & {\text{The tangent line is horizontal at the point }}\left( {4,4} \right) \cr & \cr & {\bf{b}}. \cr & {\text{Set }}f'\left( x \right) = - \frac{1}{2}{\text{ and solve for }}x \cr & \,\frac{2}{{\sqrt x }} - 1 = - \frac{1}{2} \cr & \,\frac{2}{{\sqrt x }} = 1 - \frac{1}{2} \cr & \,\frac{2}{{\sqrt x }} = \frac{1}{2} \cr & \,\,\,\,\sqrt x = 4 \cr & \,\,\,\, x = 16 \cr & {\text{Evaluate }}f\left( {16} \right) \cr & f\left( {16} \right) = 4\sqrt {16} - 16 \cr & f\left( {16} \right) = 0 \cr & {\text{The tangent line is horizontal at the point }}\left( {16,0} \right) \cr} $$