Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 76

$$\eqalign{ & {\text{Relative maximum at }}\left( {1,\frac{1}{e}} \right) \cr & {\text{Inflection points at: }}\left( {2,\frac{2}{{{e^2}}}} \right){\text{ }} \cr} $$

$$\eqalign{ & f\left( x \right) = x{e^{ - x}} \cr & {\text{Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x{e^{ - x}}} \right] \cr & f'\left( x \right) = - x{e^{ - x}} + {e^{ - x}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & - x{e^{ - x}} + {e^{ - x}} = 0 \cr & {e^{ - x}}\left( { - x + 1} \right) = 0 \cr & x = 1 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - x{e^{ - x}} + {e^{ - x}}} \right] \cr & f''\left( x \right) = x{e^{ - x}} - {e^{ - x}} - {e^{ - x}} \cr & f''\left( x \right) = x{e^{ - x}} - 2{e^{ - x}} \cr & f''\left( x \right) = \left( {x - 2} \right){e^{ - x}} \cr & \cr & {\text{Evaluate the second derivative at }}x = 1 \cr & f''\left( 1 \right) = \left( {1 - 2} \right){e^{ - 1}} < 0 \cr & {\text{Then by the second derivative test }} \cr & f\left( x \right){\text{ has a relative maximum at }}\left( {1,f\left( 1 \right)} \right) \cr & f\left( 1 \right) = \left( 1 \right){e^{ - 1}} \to {\text{Relative maximum at }}\left( {1,\frac{1}{e}} \right) \cr & \cr & {\text{Set }}f''\left( x \right) = 0 \cr & \left( {x - 2} \right){e^{ - x}} = 0 \cr & x = 2 \cr & {\text{Inflection points at: }}\left( {2,f\left( 2 \right)} \right) \cr & f\left( 2 \right) = 2{e^{ - 2}} \cr & {\text{Inflection point at: }}\left( {2,\frac{2}{{{e^2}}}} \right) \cr & \cr & {\text{Graph}} \cr} $$