Answer
$$\eqalign{
& {\text{Domain }}\left( {0,\infty } \right) \cr
& {\text{No }}y{\text{ - intercepts}} \cr
& x{\text{ - intercepts: }}\left( {1,0} \right) \cr
& {\text{Local maximum }}\left( {\frac{1}{e}, - \frac{1}{e}} \right) \cr
& {\text{No inflection points}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x\ln x \cr
& {\text{Domain }}\left( {0,\infty } \right) \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0,{\text{ }}x = 0{\text{ is not in the domain}}{\text{.}} \cr
& {\text{There are no }}y{\text{ - intercepts }} \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr
& x\ln x = 0 \cr
& x = 0{\text{ and }}x = 1 \cr
& x = 0{\text{ is not in the domain}}{\text{, then }}x = 1 \cr
& x{\text{ - intercept: }}\left( {1,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {x\ln x} \right] \cr
& f'\left( x \right) = \ln x + x\left( {\frac{1}{x}} \right) \cr
& f'\left( x \right) = \ln x + 1 \cr
& {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr
& \ln x + 1 = 0 \cr
& \ln x = - 1 \cr
& x = {e^{ - 1}} \cr
& x = \frac{1}{e} \cr
& \cr
& *{\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\ln x + 1} \right] \cr
& f''\left( x \right) = \frac{1}{x} \cr
& \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical point }}x = \frac{1}{e} \cr
& f''\left( {\frac{1}{e}} \right) = \frac{1}{{1/e}} = e> 0,{\text{ then}} \cr
& {\text{There is a local minimum at }}\left( {\frac{1}{e},f\left( {\frac{1}{e}} \right)} \right) \cr
& f\left( {\frac{1}{e}} \right) = \left( {\frac{1}{e}} \right)\ln \left( {\frac{1}{e}} \right) = - \frac{1}{e} \cr
& \to {\text{local minimum at }}\left( {\frac{1}{e}, - \frac{1}{e}} \right) \cr
& \cr
& {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr
& \frac{1}{x} = 0 \cr
& {\text{No solution}}{\text{, then there are no inflection points}}{\text{.}} \cr
& \cr
& {\text{Graph}} \cr} $$
