Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {2,0} \right) \cr
& {\text{local maximum at }}\left( {\frac{2}{7},1.9355} \right) \cr
& {\text{local minimum at }}\left( {2,0} \right) \cr
& {\text{inflection points:}} \cr
& \left( {0,0} \right),\left( { - 0.3203, - 3.6836} \right){\text{ and }}\left( {0.8918,1.1821} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{1/3}}{\left( {x - 2} \right)^2} \cr
& {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr
& \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr
& f\left( 0 \right) = {\left( 0 \right)^{1/3}}{\left( {0 - 2} \right)^2} \cr
& f\left( 0 \right) = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr
& {x^{1/3}}{\left( {x - 2} \right)^2} = 0 \cr
& {x^{1/3}} = 0,{\text{ }}x - 2 = 0 \cr
& x = 0,{\text{ }}x = 2 \cr
& x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {2,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{1/3}}{{\left( {x - 2} \right)}^2}} \right] \cr
& f'\left( x \right) = {x^{1/3}}\left[ {2\left( {x - 2} \right)} \right] + \frac{1}{3}{x^{ - 2/3}}{\left( {x - 2} \right)^2} \cr
& f'\left( x \right) = 2{x^{1/3}}\left( {x - 2} \right) + \frac{1}{3}{x^{ - 2/3}}{\left( {x - 2} \right)^2} \cr
& \cr
& {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr
& 2{x^{1/3}}\left( {x - 2} \right) + \frac{1}{3}{x^{ - 2/3}}{\left( {x - 2} \right)^2} = 0 \cr
& {x^{ - 2/3}}\left( {x - 2} \right)\left[ {2x + \frac{1}{3}\left( {x - 2} \right)} \right] = 0 \cr
& {x^{ - 2/3}}\left( {x - 2} \right)\left( {\frac{7}{3}x - \frac{2}{3}} \right) = 0 \cr
& \frac{1}{3}{x^{ - 2/3}}\left( {x - 2} \right)\left( {7x - 2} \right) = 0 \cr
& x = 2{\text{ and }}x = \frac{2}{7},{\text{ and is undefined at }}x = 0 \cr
& {\text{The domain of the function is }}\left( { - \infty ,\infty } \right),{\text{ then}} \cr
& {\text{}} \cr
& \cr
& *{\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{3}{x^{ - 2/3}}\left( {x - 2} \right)\left( {7x - 2} \right)} \right] \cr
& f''\left( x \right) = \frac{1}{3}\frac{d}{{dx}}\left[ {\left( {{x^{1/3}} - 2{x^{ - 2/3}}} \right)\left( {7x - 2} \right)} \right] \cr
& f''\left( x \right) = \frac{1}{3}\left( {{x^{1/3}} - 2{x^{ - 2/3}}} \right)\left( 7 \right) + \frac{1}{3}\left( {\frac{1}{3}{x^{ - 2/3}} + \frac{4}{3}{x^{ - 5/3}}} \right)\left( {7x - 2} \right) \cr
& {\text{Simplifying by hand and replacing}} \cr
& f''\left( x \right) = \frac{{28}}{9}{x^{1/3}} - \frac{{16}}{9}{x^{ - 2/3}} - \frac{8}{9}{x^{ - 5/3}} \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = \frac{2}{7},{\text{ }}x = 2 \cr
& *f''\left( {\frac{2}{7}} \right) = \frac{{28}}{9}{\left( {\frac{2}{7}} \right)^{1/3}} - \frac{{16}}{9}{\left( {\frac{2}{7}} \right)^{ - 2/3}} - \frac{8}{9}{\left( {\frac{2}{7}} \right)^{ - 5/3}} < 0,{\text{ then}} \cr
& {\text{There is a local maximum at }}\left( {\frac{2}{7},f\left( {\frac{2}{7}} \right)} \right) \cr
& f\left( {\frac{2}{7}} \right) = {\left( {\frac{2}{7}} \right)^{1/3}}{\left( {\frac{2}{7} - 2} \right)^2} \approx 1.9355 \cr
& \to {\text{local maximum at }}\left( {\frac{2}{7},1.9355} \right) \cr
& *f''\left( 2 \right) = \frac{{28}}{9}{\left( 2 \right)^{1/3}} - \frac{{16}}{9}{\left( 2 \right)^{ - 2/3}} - \frac{8}{9}{\left( 2 \right)^{ - 5/3}} > 0,{\text{ then}} \cr
& {\text{There is a local minimum at }}\left( {2,f\left( 2 \right)} \right) \cr
& f\left( 2 \right) = {\left( 2 \right)^{1/3}}{\left( {2 - 2} \right)^2} = 0 \cr
& \to {\text{local minimum at }}\left( {2,0} \right) \cr
& \cr
& *{\text{Use the first derivative test at the critical point}}{\text{, }}x = 0 \cr
& f'\left( { - 1} \right) = 9 > 0,{\text{ increasing}} \cr
& f'\left( 0 \right) = {\text{undefined}}{\text{,}}\;{\text{but }}x = 0\,{\text{is in the domain of }}f\left( x \right) \cr
& f'\left( {\frac{1}{{10}}} \right) \approx 3.82 > 0,{\text{ increasing}} \cr
& {\text{By the first derivative test we can conclude that there is no}} \cr
& {\text{relative extrema at }}x = 0 \cr
& \cr
& {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr
& \frac{{28}}{9}{x^{1/3}} - \frac{{16}}{9}{x^{ - 2/3}} - \frac{8}{9}{x^{ - 5/3}} = 0 \cr
& \frac{4}{9}{x^{ - 5/3}}\left( {7{x^2} - 4x - 2} \right) = 0 \cr
& {\text{Solve by using the quadratic formula}}{\text{, we obtain}} \cr
& x = \frac{{2 - 3\sqrt 7 }}{2} \approx - 0.3203 \cr
& x = \frac{{2 - 3\sqrt 7 }}{2} \approx 0.8918 \cr
& {\text{And the second derivative is undefined at }}x = 0 \cr
& f''\left( { - 0.3} \right) \approx 0.56 > 0,{\text{ concave up}} \cr
& f''\left( {\frac{1}{{10}}} \right) \approx - 48 < 0,{\text{ concave down}}{\text{, then}} \cr
& {\text{There is an inflection point at }}\left( {0,0} \right) \cr
& f\left( { - 0.3203} \right) \approx - 3.6836 \cr
& f\left( {0.8918} \right) \approx 1.1821 \cr
& {\text{The inflection points are: }} \cr
& \left( {0,0} \right),\left( { - 0.3203, - 3.6836} \right){\text{ and }}\left( {0.8918,1.1821} \right) \cr} $$
