Chapter 3 - Applications of Differentiation - 3.4 Exercises - Page 192: 34

$$\eqalign{ & {\text{Relative maximum at }}\left( {3, - 9} \right) \cr & {\text{Relative minimum at }}\left( {\frac{5}{3}, - \frac{{275}}{{27}}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = - {x^3} + 7{x^2} - 15x \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ { - {x^3} + 7{x^2} - 15x} \right] \cr & f'\left( x \right) = - 3{x^2} + 14x - 15 \cr & {\text{Set }}f'\left( x \right) = 0 \cr & - 3{x^2} + 14x - 15 = 0 \cr & 3{x^2} - 14x + 15 = 0 \cr & {\text{Factoring}} \cr & \left( {3x - 5} \right)\left( {x - 3} \right) = 0 \cr & x = \frac{5}{3},{\text{ }}x = 3 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 3{x^2} + 14x - 15} \right] \cr & f''\left( x \right) = - 6x + 14 \cr & \cr & {\text{Evaluate the second derivative at }}x = \frac{5}{3}{\text{ and }}x = 3 \cr & *f''\left( {\frac{5}{3}} \right) = - 6\left( {\frac{5}{3}} \right) + 14 = 4 > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( {\frac{5}{3},f\left( {\frac{5}{3}} \right)} \right) \cr & f\left( {\frac{5}{3}} \right) = - {\left( {\frac{5}{3}} \right)^3} + 7{\left( {\frac{5}{3}} \right)^2} - 15\left( {\frac{5}{3}} \right) = - \frac{{275}}{{27}} \cr & {\text{Relative minimum at }}\left( {\frac{5}{3}, - \frac{{275}}{{27}}} \right) \cr & *f''\left( 3 \right) = - 6\left( 3 \right) + 14 = - 4 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative maximum at }}\left( {3,f\left( 3 \right)} \right) \cr & f\left( 3 \right) = - {\left( 3 \right)^3} + 7{\left( 3 \right)^2} - 15\left( 3 \right) = - 9 \cr & {\text{Relative maximum at }}\left( {3, - 9} \right) \cr} $$