Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 32

$$\eqalign{ & {\text{No }}y{\text{ - intercepts}} \cr & x{\text{ - intercepts: }}\left( {1,0} \right) \cr & {\text{Local maximum }}\left( {\sqrt e ,\frac{1}{{2e}}} \right) \cr & {\text{Inflection point }}\left( {{e^{5/6}},\frac{5}{{6{e^{5/3}}}}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{\ln x}}{{{x^2}}} \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \frac{{\ln \left( 0 \right)}}{{{{\left( 0 \right)}^2}}},{\text{ Inderteminate}}{\text{, then }} \cr & {\text{There are no }}y{\text{ - intercepts }} \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = \frac{{\ln x}}{{{x^2}}} \cr & \ln x = 0 \cr & 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[ {\frac{{\ln x}}{{{x^2}}}} \right] \cr & f'\left( x \right) = \frac{{{x^2}\left( {\frac{1}{x}} \right) - \ln x\left( {2x} \right)}}{{{{\left( {{x^2}} \right)}^2}}} \cr & f'\left( x \right) = \frac{{x - 2x\ln x}}{{{x^4}}} \cr & f'\left( x \right) = \frac{{1 - 2\ln x}}{{{x^3}}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \frac{{1 - 2\ln x}}{{{x^3}}} = 0 \cr & 1 - 2\ln x = 0 \cr & \ln x = \frac{1}{2} \cr & x = {e^{1/2}} \cr & x = \sqrt e \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{1 - 2\ln x}}{{{x^3}}}} \right] \cr & f''\left( x \right) = \frac{{{x^3}\left( { - \frac{2}{x}} \right) - \left( {1 - 2\ln x} \right)\left( {3{x^2}} \right)}}{{{x^6}}} \cr & f''\left( x \right) = \frac{{ - 2{x^2} - 3{x^2}\left( {1 - 2\ln x} \right)}}{{{x^6}}} \cr & f''\left( x \right) = \frac{{ - 2 - 3\left( {1 - 2\ln x} \right)}}{{{x^4}}} \cr & f''\left( x \right) = \frac{{ - 5 + 6\ln x}}{{{x^4}}} \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical point }}x = \sqrt e \cr & f''\left( {\sqrt e } \right) = \frac{{ - 5 + 6\ln \left( {\sqrt e } \right)}}{{{{\left( {\sqrt e } \right)}^4}}} = - \frac{2}{{{e^2}}} < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}\left( {\sqrt e ,f\left( {\sqrt e } \right)} \right) \cr & f\left( {\sqrt e } \right) = \frac{{\ln \left( {\sqrt e } \right)}}{{{{\left( {\sqrt e } \right)}^2}}} = \frac{1}{{2e}} \cr & \to {\text{local maximum at }}\left( {\sqrt e ,\frac{1}{{2e}}} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{{ - 5 + 6\ln x}}{{{x^4}}} = 0 \cr & - 5 + 6\ln x = 0 \cr & \ln x = \frac{5}{6} \cr & x = {e^{5/6}} \cr & {\text{Inflection point: }}\left( {{e^{5/6}},f\left( {{e^{5/6}}} \right)} \right) \cr & f\left( {{e^{5/6}}} \right) = \frac{{\ln \left( {{e^{5/6}}} \right)}}{{{{\left( {{e^{5/6}}} \right)}^2}}} = \frac{5}{{6{e^{5/3}}}} \cr & {\text{The inflection point is: }}\left( {{e^{5/6}},\frac{5}{{6{e^{5/3}}}}} \right) \cr & \cr & {\text{Graph}} \cr} $$