Chapter 3 - The Derivative In Graphing And Applications - 3.3 Analysis Of Functions III: Rational Functions, Cusps, And Vertical Tangents - Exercises Set 3.3 - Page 215: 38

$$\eqalign{ & {\text{The critical points is: }}\left( {4,2} \right) \cr & {\text{The inflection points is: }}\left( {\frac{4}{9}, - 6} \right) \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{{8\left( {\sqrt x - 1} \right)}}{x} \cr & {\text{Simplify the function}} \cr & f\left( x \right) = \frac{{8\sqrt x }}{x} - \frac{8}{x} \cr & f\left( x \right) = \frac{8}{{\sqrt x }} - \frac{8}{x} \cr & f'\left( x \right) = - 4{x^{ - 3/2}} + \frac{8}{{{x^2}}} \cr & {\text{Calculate the critical points}} \cr & f'\left( x \right) = 0 \cr & - 4{x^{ - 3/2}} + 8{x^{ - 2}} = 0 \cr & - 4{x^{ - 2}}\left( {{x^{1/2}} - 2} \right) = 0 \cr & {\text{Solving we obtain}} \cr & x = 4 \cr & {\text{The derivative is not defined at }}x = 0 \cr & {\text{Calculate }}f\left( 0 \right){\text{ and }}f\left( 4 \right) \cr & f\left( 4 \right) = 2 \cr & {\text{The critical points is: }}\left( {4,2} \right) \cr & \cr & {\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 4{x^{ - 3/2}} + 8{x^{ - 2}}} \right] \cr & f''\left( x \right) = - 4\left( { - \frac{3}{2}} \right){x^{ - 5/2}} + 8\left( { - 2} \right){x^{ - 3}} \cr & f''\left( x \right) = 6{x^{ - 5/2}} - 4{x^{ - 3}} \cr & 6{x^{ - 5/2}} - 4{x^{ - 3}} = 0 \cr & 2{x^{ - 3}}\left( {3{x^{1/2}} - 2} \right) = 0 \cr & {\text{Solving we obtain}} \cr & x = \frac{4}{9},{\text{ and it's not defined at }}x = 0 \cr & f\left( {\frac{4}{9}} \right) = - 6 \cr & {\text{The inflection points is: }}\left( {\frac{4}{9}, - 6} \right) \cr} $$