Chapter 3 - The Derivative In Graphing And Applications - Chapter 3 Review Exercises - Page 260: 34

$$\eqalign{ & {\text{relative minimum of 6 at }}x = - 1{\text{ and }}x = 1 \cr & {\text{relative maximum of 8 at }}x = 0 \cr} $$

$$\eqalign{ & f\left( x \right) = {x^4} - 2{x^2} + 7 \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^4} - 2{x^2} + 7} \right] \cr & f'\left( x \right) = 4{x^3} - 4x \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & 4{x^3} - 4x = 0 \cr & 4x\left( {{x^2} - 1} \right) = 0 \cr & {\text{The critical points are }}x = 0,\,\,\,x = - 1{\text{ and }}x = 1 \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {4{x^3} - 4x} \right] \cr & f''\left( x \right) = 12{x^2} - 4 \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}f''\left( 0 \right) \cr & f''\left( 0 \right) = 12{\left( 0 \right)^2} - 4 \cr & f''\left( 0 \right) = - 4 \cr & f''\left( 0 \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local maximum at }}x = 0 \cr & \cr & {\text{Evaluate }}f''\left( { - 1} \right) \cr & f''\left( { - 1} \right) = 12{\left( { - 1} \right)^2} - 4 \cr & f''\left( { - 1} \right) = 8 \cr & f''\left( { - 1} \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local minimum at }}x = - 1 \cr & \cr & {\text{Evaluate }}f''\left( 1 \right) \cr & f''\left( 1 \right) = 12{\left( 1 \right)^2} - 4 \cr & f''\left( 1 \right) = 8 \cr & f''\left( 1 \right) > 0,{\text{ then }}f\left( x \right){\text{ has a relative minimum at }}x = 1 \cr & f\left( 1 \right) = {\left( 1 \right)^4} - 2{\left( 1 \right)^2} + 7 = 6 \cr} $$