Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& x{\text{ - intercepts: }}\left( {0,0} \right),\left( { \pm 3\sqrt 3 ,0} \right) \cr
& {\text{Critical points }}x = - 1,{\text{ }}x = 0{\text{ and }}x = 1 \cr
& \left( { - 1,2} \right){\text{ Local maximum}} \cr
& \left( {1, - 2} \right){\text{ Local minimum}} \cr
& \left( {0,0} \right){\text{ Inflection point}} \cr
& {\text{No vertical asymptotes}} \cr
& {\text{No horizontal asymptotes}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x - 3{x^{1/3}} \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) - 3{\left( 0 \right)^{1/3}} \cr
& f\left( 0 \right) = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr
& x - 3{x^{1/3}} = 0 \cr
& x\left( {1 - 3{x^{ - 2/3}}} \right) = 0 \cr
& x = 0,{\text{ }}1 - 3{x^{ - 2/3}} = 0 \to x = \pm 3\sqrt 3 \cr
& x{\text{ - intercepts: }}\left( {0,0} \right),\left( { \pm 3\sqrt 3 ,0} \right) \cr
& \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {x - 3{x^{1/3}}} \right] \cr
& f'\left( x \right) = 1 - {x^{ - 2/3}} \cr
& {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr
& 1 - {x^{ - 2/3}} = 0 \cr
& {x^{ - 2/3}}\left( {{x^{2/3}} - 1} \right) = 0 \cr
& {x^{ - 2/3}} = 0,{\text{ }}{x^{2/3}} - 1 = 0 \cr
& {\text{The derivate is undefined at }}x = 0 \cr
& {\text{ }}{x^{2/3}} - 1 = 0 \cr
& x = \pm 1 \cr
& {\text{The critical points are }}x = 0,{\text{ }}x = - 1{\text{ and }}x = 1 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {1 - {x^{ - 2/3}}} \right] \cr
& f''\left( x \right) = \frac{2}{3}{x^{ - 5/3}} \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = \pm 1 \cr
& \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical point }}x = 0 \cr
& f''\left( 0 \right) = \frac{2}{3}{\left( 0 \right)^{ - 5/3}} = 0,{\text{The second derivative test is inconclusive}} \cr
& {\text{using the first derivative test:}} \cr
& f'\left( { - 0.5} \right) = 1 - {\left( { - 0.5} \right)^{ - 2/3}} = - 0.58 < 0,{\text{ decreasing}} \cr
& f'\left( 0 \right) = 1 - {\left( 0 \right)^{ - 2/3}}{\text{, undefined but }}x = 0\,{\text{is in the domain of }}f \cr
& f'\left( {0.5} \right) = 1 - {\left( {0.5} \right)^{ - 2/3}} = - 0.58 < 0,{\text{ decreasing}} \cr
& {\text{By the first derivative test we can conclude that there is no}} \cr
& {\text{relative extrema at }}x = 0 \cr
& \cr
& f''\left( { - 1} \right) = \frac{2}{3}{\left( { - 1} \right)^{ - 5/3}} = - \frac{2}{3} < 0 \to {\text{local maximum at }}\left( {1,f\left( 1 \right)} \right) \cr
& \left( { - 1,f\left( 1 \right)} \right) \to \left( { - 1,2} \right){\text{ Local maximum}} \cr
& f''\left( 1 \right) = \frac{2}{3}{\left( 1 \right)^{ - 5/3}} = \frac{2}{3} > 0 \to {\text{local minimum at }}\left( {1,f\left( 1 \right)} \right) \cr
& \left( {1,f\left( 1 \right)} \right) \to \left( {1, - 2} \right){\text{ Local minimum}} \cr
& \cr
& {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr
& \frac{2}{3}{x^{ - 5/3}} = 0 \cr
& {\text{The second derivative is not defined at }}x = 0 \cr
& f''\left( { - 1} \right) < 0,{\text{ }}\left( {{\text{Concave down}}} \right) \cr
& f''\left( 1 \right) > 0,{\text{ }}\left( {{\text{Concave up}}} \right) \cr
& x = 0{\text{ is in the domain of the function}}{\text{, then}} \cr
& f\left( 0 \right) = 0 \cr
& \left( {0,0} \right){\text{ Is an inflection point}} \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {x - 3{x^{1/3}}} \right) = + \infty \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {x - 3{x^{1/3}}} \right) = - \infty \cr
& {\text{No horizontal asymptotes}} \cr
& \cr
& {\text{Graph}} \cr} $$
