Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 78

$$\eqalign{ & {\text{Relative maximum at }}\left( {\frac{5}{3},96.942} \right) \cr & {\text{Inflection points at: }}\left( {\frac{4}{3},70.79753} \right){\text{ }} \cr} $$

$$\eqalign{ & f\left( x \right) = - 2 + {e^{3x}}\left( {4 - 2x} \right) \cr & {\text{Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ { - 2 + {e^{3x}}\left( {4 - 2x} \right)} \right] \cr & f'\left( x \right) = 3{e^{3x}}\left( {4 - 2x} \right) - 2{e^{3x}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & 3{e^{3x}}\left( {4 - 2x} \right) - 2{e^{3x}} = 0 \cr & \left[ {3\left( {4 - 2x} \right) - 2} \right]{e^{3x}} = 0 \cr & 12 - 6x - 2 = 0 \cr & 10 - 6x = 0 \cr & x = \frac{5}{3} \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {3{e^{3x}}\left( {4 - 2x} \right) - 2{e^{3x}}} \right] \cr & f''\left( x \right) = - 6{e^{3x}} + 9\left( {4 - 2x} \right){e^{3x}} - 6{e^{3x}} \cr & \cr & {\text{Evaluate the second derivative at }}x = \frac{5}{3} \cr & f''\left( {\frac{5}{3}} \right) = - 6{e^{3\left( {5/3} \right)}} + 9\left( {4 - 2\left( {\frac{5}{3}} \right)} \right){e^{3\left( {5/3} \right)}} - 6{e^{3\left( {5/3} \right)}} < 0 \cr & {\text{Then by the second derivative test }} \cr & f\left( x \right){\text{ has a relative maximum at }}\left( {\frac{5}{3},f\left( {\frac{5}{3}} \right)} \right) \cr & f\left( {\frac{5}{3}} \right) = - 2 + {e^{3\left( {5/3} \right)}}\left( {4 - 2\left( {\frac{5}{3}} \right)} \right) \approx 96.942 \cr & \to {\text{Relative maximum at }}\left( {\frac{5}{3},96.942} \right) \cr & \cr & {\text{Set }}f''\left( x \right) = 0 \cr & - 6{e^{3x}} + 9\left( {4 - 2x} \right){e^{3x}} - 6{e^{3x}} = 0 \cr & 3\left[ { - 2 + 3\left( {4 - 2x} \right) - 2} \right]{e^{3x}} = 0 \cr & 3\left( { - 2 + 12 - 6x - 2} \right){e^{3x}} = 0 \cr & 3\left( {8 - 6x} \right){e^{3x}} = 0 \cr & x = \frac{8}{6} = \frac{4}{3} \cr & {\text{Inflection points at: }}\left( {\frac{4}{3},f\left( {\frac{4}{3}} \right)} \right) \cr & f\left( 2 \right) = 2{e^{ - 2}} \cr & {\text{Inflection point at: }}\left( {2,\frac{2}{{{e^2}}}} \right) \cr & f\left( {\frac{4}{3}} \right) = - 2 + {e^{3\left( {4/3} \right)}}\left( {4 - 2\left( {\frac{4}{3}} \right)} \right) \approx 70.79753 \cr & {\text{Graph}} \cr} $$