Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 77

$$\eqalign{ & {\text{Relative maximum at }}\left( { - 1,1 + e} \right) \cr & {\text{Inflection points at: }}\left( {0,3} \right) \cr} $$

$$\eqalign{ & g\left( t \right) = 1 + \left( {2 + t} \right){e^{ - t}} \cr & {\text{Calculate the first derivative}} \cr & g'\left( t \right) = \frac{d}{{dt}}\left[ {1 + \left( {2 + t} \right){e^{ - t}}} \right] \cr & g'\left( t \right) = - \left( {2 + t} \right){e^{ - t}} + {e^{ - t}} \cr & g'\left( t \right) = - 2{e^{ - t}} - t{e^{ - t}} + {e^{ - t}} \cr & g'\left( t \right) = - t{e^{ - t}} - {e^{ - t}} \cr & {\text{Set }}g'\left( t \right) = 0 \cr & - t{e^{ - t}} - {e^{ - t}} = 0 \cr & {e^{ - t}}\left( { - t - 1} \right) = 0 \cr & t = - 1 \cr & \cr & *{\text{Calculate the second derivative}} \cr & g''\left( t \right) = \frac{d}{{dt}}\left[ { - t{e^{ - t}} - {e^{ - t}}} \right] \cr & g''\left( t \right) = t{e^{ - t}} - {e^{ - t}} + {e^{ - t}} \cr & g''\left( t \right) = t{e^{ - t}} \cr & \cr & {\text{Evaluate the second derivative at }}t = - 1 \cr & g''\left( { - 1} \right) = \left( { - 1} \right){e^{ - 1}} < 0 \cr & {\text{Then by the second derivative test }} \cr & g\left( t \right){\text{ has a relative maximum at }}\left( { - 1,g\left( { - 1} \right)} \right) \cr & g\left( 1 \right) = 1 + \left( {2 - 1} \right){e^{ - \left( { - 1} \right)}} = 1 + e \to \left( {1,1 + e} \right) \cr & \cr & {\text{Set }}g''\left( t \right) = 0 \cr & t{e^{ - t}} = 0 \cr & t = 0 \cr & {\text{Inflection points at: }}\left( {0,g\left( 0 \right)} \right) \cr & g\left( 0 \right) = 1 + \left( {2 + 0} \right){e^{ - 0}} = 3 \cr & {\text{Inflection point at: }}\left( {0,3} \right) \cr & \cr & {\text{Graph}} \cr} $$