Chapter 3 - Applications of Differentiation - Review Exercises - Page 238: 42

$$\eqalign{ & {\text{Relative maximum at }}\left( { - 4,68} \right) \cr & {\text{Relative minimum at }}\left( {\frac{1}{3}, - \frac{{361}}{{27}}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = 2{x^3} + 11{x^2} - 8x - 12 \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3} + 11{x^2} - 8x - 12} \right] \cr & f'\left( x \right) = 6{x^2} + 22x - 8 \cr & {\text{Set }}f'\left( x \right) = 0 \cr & 6{x^2} + 22x - 8 = 0 \cr & 2\left( {3{x^2} + 11x - 4} \right) = 0 \cr & 2\left( {3x - 1} \right)\left( {x + 4} \right) = 0 \cr & x = \frac{1}{3},{\text{ }}x = - 4 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {6{x^2} + 22x - 8} \right] \cr & f'\left( x \right) = 12x + 22 \cr & \cr & {\text{Evaluate the second derivative at }}x = \frac{1}{3}{\text{ and }}x = - 4 \cr & *f''\left( { - 4} \right) = 12\left( { - 4} \right) + 22 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative maximum at }}\left( { - 4,f\left( { - 4} \right)} \right) \cr & f\left( { - 4} \right) = 2{\left( { - 4} \right)^3} + 11{\left( { - 4} \right)^2} - 8\left( { - 4} \right) - 12 = 68 \cr & {\text{Relative maximum at }}\left( { - 4,68} \right) \cr & *f''\left( {\frac{1}{3}} \right) = 12\left( {\frac{1}{3}} \right) + 22 > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( {\frac{1}{3},f\left( {\frac{1}{3}} \right)} \right) \cr & f\left( {\frac{1}{3}} \right) = 2{\left( {\frac{1}{3}} \right)^3} + 11{\left( {\frac{1}{3}} \right)^2} - 8\left( {\frac{1}{3}} \right) - 12 = - \frac{{361}}{{27}} \cr & {\text{Relative minimum at }}\left( {\frac{1}{3}, - \frac{{361}}{{27}}} \right) \cr} $$