Chapter 3 - Applications of Differentiation - Review Exercises - Page 238: 43

$$\eqalign{ & {\text{Relative maximum at }}\left( { \pm \frac{1}{{\sqrt 2 }},\frac{1}{2}} \right) \cr & {\text{Relative minimum at }}\left( {0,0} \right) \cr} $$

$$\eqalign{ & g\left( x \right) = 2{x^2}\left( {1 - {x^2}} \right) \cr & g\left( x \right) = 2{x^2} - 2{x^4} \cr & {\text{*Calculate the first derivative}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^2} - 2{x^4}} \right] \cr & g'\left( x \right) = 4x - 8{x^3} \cr & {\text{Set }}g'\left( x \right) = 0 \cr & 4x - 8{x^3} = 0 \cr & 4x\left( {1 - 2{x^2}} \right) = 0 \cr & x = 0,{\text{ }}x = - \frac{1}{{\sqrt 2 }},{\text{ }}x = \frac{1}{{\sqrt 2 }} \cr & \cr & *{\text{Calculate the second derivative}} \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ {4x - 8{x^3}} \right] \cr & g''\left( x \right) = 4 - 24{x^2} \cr & \cr & {\text{Evaluate the second derivative at }} \cr & x = 0,{\text{ }}x = - \frac{1}{{\sqrt 2 }},{\text{ }}x = \frac{1}{{\sqrt 2 }} \cr & *g''\left( 0 \right) = 4 - 24{\left( 0 \right)^2} = 4 > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & g\left( x \right){\text{ has a relative minimum at }}\left( {0,g\left( 0 \right)} \right) \cr & g\left( 0 \right) = 2{\left( 0 \right)^2}\left( {1 - {{\left( 0 \right)}^2}} \right) = 0 \to \left( {0,0} \right) \cr & \cr & *g''\left( { - \frac{1}{{\sqrt 2 }}} \right) = 4 - 24{\left( { - \frac{1}{{\sqrt 2 }}} \right)^2} = - 8 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & g\left( x \right){\text{ has a relative maximum at }}\left( { - \frac{1}{{\sqrt 2 }},g\left( { - \frac{1}{{\sqrt 2 }}} \right)} \right) \cr & g\left( { - \frac{1}{{\sqrt 2 }}} \right) = 2{\left( { - \frac{1}{{\sqrt 2 }}} \right)^2}\left( {1 - {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2}} \right) = \frac{1}{2} \to \left( { - \frac{1}{{\sqrt 2 }},\frac{1}{2}} \right) \cr & {\text{Relative maximum at }}\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{2}} \right) \cr & *g''\left( {\frac{1}{{\sqrt 2 }}} \right) = 4 - 24{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = - 8 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & g\left( x \right){\text{ has a relative maximum at }}\left( { - \frac{1}{{\sqrt 2 }},g\left( { - \frac{1}{{\sqrt 2 }}} \right)} \right) \cr & g\left( {\frac{1}{{\sqrt 2 }}} \right) = 2{\left( {\frac{1}{{\sqrt 2 }}} \right)^2}\left( {1 - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right) = \frac{1}{2} \to \left( {\frac{1}{{\sqrt 2 }},\frac{1}{2}} \right) \cr & {\text{Relative maximum at }}\left( {\frac{1}{{\sqrt 2 }},\frac{1}{2}} \right) \cr} $$