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: 35

$$\eqalign{ & {\text{critical points: }}\left( {0,0} \right),{\text{ }}\left( {1,3} \right) \cr & {\text{inflection points: }}\left( { - 2, - 6\root 3 \of 2 } \right){\text{ and }}\left( {0,0} \right) \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = 4{x^{1/3}} - {x^{4/3}} \cr & {\text{Differentiate }}f'\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {4{x^{1/3}} - {x^{4/3}}} \right] \cr & f'\left( x \right) = \frac{4}{3}{x^{ - 2/3}} - \frac{4}{3}{x^{1/3}} \cr & {\text{Calculate the critical points set the derivative equal to 0}} \cr & \frac{4}{3}{x^{ - 2/3}} - \frac{4}{3}{x^{1/3}} = 0 \cr & \frac{4}{3}{x^{ - 2/3}}\left( {1 - x} \right) = 0 \cr & \frac{4}{3}{x^{ - 2/3}} = 0,{\text{ }}1 - x = 0 \cr & x = 1 \cr & {\text{The derivative is not defined at }}x = 0 \cr & {\text{Calculate }}f\left( 0 \right),{\text{ }}f\left( 1 \right) \cr & f\left( 0 \right) = 0,{\text{ }}f\left( 1 \right) = 3 \cr & {\text{Therefore, the critical points are:}} \cr & \left( {0,0} \right),{\text{ }}\left( {1,3} \right) \cr & \cr & {\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{4}{3}{x^{ - 2/3}} - \frac{4}{3}{x^{1/3}}} \right] \cr & f''\left( x \right) = - \frac{8}{9}{x^{ - 5/3}} - \frac{4}{9}{x^{ - 2/3}} \cr & \cr & {\text{Set }}f''\left( x \right) = 0 \cr & - \frac{8}{9}{x^{ - 5/3}} - \frac{4}{9}{x^{ - 2/3}} = 0 \cr & - \frac{4}{9}{x^{ - 5/3}}\left( {2 + x} \right) = 0 \cr & 2 + x = 0 \cr & x = - 2 \cr & {\text{And the second derivative is not defined at }}x = 0 \cr & {\text{Calculate }}f\left( { - 2} \right){\text{ and }}f\left( 0 \right) \cr & f\left( { - 2} \right) = - 6\root 3 \of 2 \cr & f\left( 0 \right) = 0 \cr & {\text{Therefore, the inflection points are:}} \cr & \left( { - 2, - 6\root 3 \of 2 } \right){\text{ and }}\left( {0,0} \right) \cr} $$