Chapter 3 - Applications of Differentiation - 3.4 Exercises - Page 192: 36

$$\eqalign{ & {\text{Relative maximum at }}\left( { - 1,3} \right) \cr & {\text{Relative minimum at }}\left( {0,0} \right) \cr & {\text{Relative maximum at }}\left( {4,128} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = - {x^4} + 4{x^3} + 8{x^2} \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ { - {x^4} + 4{x^3} + 8{x^2}} \right] \cr & f'\left( x \right) = - 4{x^3} + 12{x^2} + 16x \cr & {\text{Set }}f'\left( x \right) = 0 \cr & - 4{x^3} + 12{x^2} + 16x = 0 \cr & {\text{Factoring}} \cr & - 4x\left( {{x^2} - 3x - 4} \right) = 0 \cr & - 4x\left( {x - 4} \right)\left( {x + 1} \right) = 0 \cr & x = - 1,{\text{ }}x = 0,{\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[ { - 4{x^3} + 12{x^2} + 16x} \right] \cr & f''\left( x \right) = - 12{x^2} + 24x + 16 \cr & {\text{Evaluate the second derivative at }}x = - 1,{\text{ }}x = 0,{\text{ }}x = 4 \cr & *f''\left( { - 1} \right) = - 12{\left( { - 1} \right)^2} + 24\left( { - 1} \right) + 16 = - 20 \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( { - 1,f\left( { - 1} \right)} \right) \cr & f\left( { - 1} \right) = - {\left( { - 1} \right)^4} + 4{\left( { - 1} \right)^3} + 8{\left( { - 1} \right)^2} = 3 \cr & {\text{Relative maximum at }}\left( { - 1,3} \right) \cr & *f''\left( 0 \right) = - 12{\left( 0 \right)^2} + 24\left( 0 \right) + 16 = 16 > 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( {0,f\left( 0 \right)} \right) \cr & f\left( 0 \right) = - {\left( 0 \right)^4} + 4{\left( 0 \right)^3} + 8{\left( 0 \right)^2} = 0 \cr & {\text{Relative minimum at }}\left( {0,0} \right) \cr & *f''\left( 4 \right) = - 12{\left( 4 \right)^2} + 24\left( 4 \right) + 16 < 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) = - {\left( 4 \right)^4} + 4{\left( 4 \right)^3} + 8{\left( 4 \right)^2} = 128 \cr & {\text{Relative maximum at }}\left( {4,128} \right) \cr} $$