Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 30

$$\eqalign{ & {\text{no }}y{\text{ - intercepts}} \cr & x{\text{ - intercept: }}\left( {1,0} \right) \cr & {\text{Local minimum at }}\left( {\frac{1}{{\sqrt e }}, - \frac{1}{{2e}}} \right) \cr & {\text{Inflection points: }}\left( {\frac{1}{{e\sqrt e }}, - \frac{3}{{2{e^3}}}} \right) \cr & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & g\left( x \right) = {x^2}\ln x \cr & {\text{Domain }}\left( {0,\infty } \right) \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0,\,x = 0{\text{ is not in the domain}}{\text{, so}} \cr & {\text{There are no }}y{\text{ - intercepts}} \cr & \cr & {\text{Find the }}x{\text{ intercepts}}{\text{, let }}g\left( x \right) = 0 \cr & {x^2}\ln x = 0 \cr & {x^2} = 0,{\text{ }}\ln x = 0 \cr & x = 0,{\text{ }}x = 1 \cr & x = 0{\text{ is not in the domain then}}{\text{, }}x = 1 \cr & x{\text{ - intercept: }}\left( {1,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2}\ln x} \right] \cr & g'\left( x \right) = {x^2}\left( {\frac{1}{x}} \right) + \ln x\left( {2x} \right) \cr & g'\left( x \right) = x + 2x\ln x \cr & \cr & {\text{Let }}g'\left( x \right) = 0{\text{ to find critical points}} \cr & x\left( {1 + 2\ln x} \right) = 0 \cr & x = 0,{\text{ }}1 + 2\ln x = 0 \to x = {e^{ - 1/2}} = \frac{1}{{\sqrt e }} \cr & x = 0{\text{ is not in the domain then}}{\text{, the only critical value is}} \cr & x = \frac{1}{{\sqrt e }} \cr & \cr & *{\text{Find the second derivative}} \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ {x + 2x\ln x} \right] \cr & g''\left( x \right) = 1 + 2x\left( {\frac{1}{x}} \right) + \ln x\left( 2 \right) \cr & g''\left( x \right) = 1 + 2 + \ln x\left( 2 \right) \cr & g''\left( x \right) = 3 + 2\ln x \cr & \cr & {\text{Evaluate }}g''\left( x \right){\text{ at the critical point }}x = \frac{1}{{\sqrt e }} \cr & g''\left( {\frac{1}{{\sqrt e }}} \right) = 3 + 2\ln \left( {\frac{1}{{\sqrt e }}} \right) = 2 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}\left( {\frac{1}{{\sqrt e }},g\left( {\frac{1}{{\sqrt e }}} \right)} \right) \cr & g\left( {\frac{1}{{\sqrt e }}} \right) = {\left( {\frac{1}{{\sqrt e }}} \right)^2}\ln \left( {\frac{1}{{\sqrt e }}} \right) = \frac{1}{e}\left( { - \frac{1}{2}} \right) = - \frac{1}{{2e}} \cr & \to {\text{local minimum at }}\left( {\frac{1}{{\sqrt e }}, - \frac{1}{{2e}}} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}g''\left( x \right) = 0 \cr & 3 + 2\ln x = 0 \cr & \ln x = - \frac{3}{2} \cr & x = {e^{ - \frac{3}{2}}} \cr & x = \frac{1}{{e\sqrt e }} \cr & g\left( {\frac{1}{{e\sqrt e }}} \right) = {\left( {\frac{1}{{e\sqrt e }}} \right)^2}\ln \left( {\frac{1}{{e\sqrt e }}} \right) = \frac{1}{{{e^3}}}\left( { - \frac{3}{2}} \right) = - \frac{3}{{2{e^3}}} \cr & {\text{The inflection point is at }}\left( {\frac{1}{{e\sqrt e }}, - \frac{3}{{2{e^3}}}} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$