Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 80

$$\eqalign{ & {\text{Critical points: }}x = 0,\,\,x = - \frac{1}{{\sqrt 3 }}{\text{ and }}x = \frac{1}{{\sqrt 3 }} \cr & {\text{local minimum at }}x = 0 \cr & {\text{local maximum at }}x = - \frac{1}{{\sqrt 3 }}{\text{ and }}x = \frac{1}{{\sqrt 3 }} \cr & \cr} $$

$$\eqalign{ & g\left( x \right) = \frac{{{x^4}}}{{2 - 12{x^2}}} \cr & {\text{Calculate the first derivative }} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{{2 - 12{x^2}}}} \right] \cr & g'\left( x \right) = \frac{{4{x^3}\left( {2 - 12{x^2}} \right) - {x^4}\left( { - 24x} \right)}}{{{{\left( {2 - 12{x^2}} \right)}^2}}} \cr & g'\left( x \right) = \frac{{8{x^3} - 48{x^5} + 24{x^5}}}{{{{\left( {2 - 12{x^2}} \right)}^2}}} \cr & g'\left( x \right) = \frac{{8{x^3} - 24{x^5}}}{{{{\left( {2 - 12{x^2}} \right)}^2}}} \cr & {\text{Find the critical points}}{\text{, set }}g'\left( x \right) = 0 \cr & \frac{{8{x^3} - 24{x^5}}}{{{{\left( {2 - 12{x^2}} \right)}^2}}} = 0 \cr & 8{x^3} - 24{x^5} = 0 \cr & 8{x^3}\left( {1 - 3{x^2}} \right) = 0 \cr & {\text{The critical points are }}x = 0,{\text{ }}x = \frac{1}{{\sqrt 3 }}{\text{ and }}x = - \frac{1}{{\sqrt 3 }} \cr & \cr & {\text{Calculate the second derivative }} \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{8{x^3} - 24{x^5}}}{{{{\left( {2 - 12{x^2}} \right)}^2}}}} \right] \cr & g''\left( x \right) = \frac{{{{\left( {2 - 12{x^2}} \right)}^2}\left( {24{x^2} - 120{x^4}} \right) - 2\left( {8{x^3} - 24{x^5}} \right)\left( {2 - 12{x^2}} \right)\left( { - 24x} \right)}}{{{{\left( {2 - 12{x^2}} \right)}^4}}} \cr & g''\left( x \right) = \frac{{\left( {2 - 12{x^2}} \right)\left( {24{x^2} - 120{x^4}} \right) + 48x\left( {8{x^3} - 24{x^5}} \right)}}{{{{\left( {2 - 12{x^2}} \right)}^3}}} \cr & g''\left( x \right) = \frac{{\left( {1 - 6{x^2}} \right)\left( {12{x^2} - 60{x^4}} \right) + 24x\left( {4{x^3} - 12{x^5}} \right)}}{{2{{\left( {1 - 6{x^2}} \right)}^3}}} \cr & {\text{simplifying}} \cr & g''\left( x \right) = \frac{{ - 36{x^6} + 18{x^4} - 6{x^2}}}{{{{\left( {6{x^2} - 1} \right)}^3}}} \cr & \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}g''\left( { - \frac{1}{{\sqrt 3 }}} \right) \cr & g''\left( { - 1/\sqrt 3 } \right) = \frac{{ - 36{{\left( { - 1/\sqrt 3 } \right)}^6} + 18{{\left( { - 1/\sqrt 3 } \right)}^4} - 6{{\left( { - 1/\sqrt 3 } \right)}^2}}}{{{{\left( {6{{\left( { - 1/\sqrt 3 } \right)}^2} - 1} \right)}^3}}} \cr & g''\left( { - 1/\sqrt 3 } \right) = - \frac{4}{3} \cr & g''\left( { - \frac{1}{{\sqrt 3 }}} \right) < 0,{\text{ then }}g\left( x \right){\text{ has a local maximum at }}x = - \frac{1}{{\sqrt 3 }} \cr & \cr & {\text{Evaluate }}g''\left( 0 \right) \cr & g''\left( 0 \right) = \frac{{ - 36{{\left( 0 \right)}^6} + 18{{\left( 0 \right)}^4} - 6{{\left( 0 \right)}^2}}}{{{{\left( {6{{\left( 0 \right)}^2} - 1} \right)}^3}}} \cr & g''\left( 0 \right) = 0,{\text{ then the test is inconclusive; may have a local maximum}}{\text{,}} \cr & {\text{local minimum}}{\text{, or neither at 0}}{\text{.}} \cr & {\text{Evaluating }}g'\left( { - 0.1} \right){\text{ and }}g'\left( {0.1} \right) \cr & g'\left( { - 0.1} \right) = \frac{{8{{\left( { - 0.1} \right)}^3} - 24{{\left( { - 0.1} \right)}^5}}}{{{{\left( {2 - 12{{\left( { - 0.1} \right)}^2}} \right)}^2}}} \approx - 0.00219 \cr & g'\left( {0.1} \right) = \frac{{8{{\left( {0.1} \right)}^3} - 24{{\left( {0.1} \right)}^5}}}{{{{\left( {2 - 12{{\left( {0.1} \right)}^2}} \right)}^2}}} \approx 0.00219 \cr & {\text{The first derivative changes negative to positive}}{\text{, then }}p\left( x \right){\text{ has a}} \cr & {\text{local minimun at }}x = 0. \cr & \cr & {\text{Evaluate }}g''\left( {\frac{1}{{\sqrt 3 }}} \right) \cr & g''\left( {1/\sqrt 3 } \right) = \frac{{ - 36{{\left( {1/\sqrt 3 } \right)}^6} + 18{{\left( {1/\sqrt 3 } \right)}^4} - 6{{\left( {1/\sqrt 3 } \right)}^2}}}{{{{\left( {6{{\left( {1/\sqrt 3 } \right)}^2} - 1} \right)}^3}}} \cr & g''\left( {1/\sqrt 3 } \right) = - \frac{4}{3} \cr & g''\left( { - \frac{1}{{\sqrt 3 }}} \right) < 0,{\text{ then }}g\left( x \right){\text{ has a local maximum at }}x = - \frac{1}{{\sqrt 3 }} \cr & \cr & {\text{Critical points: }}x = 0,\,\,x = - \frac{1}{{\sqrt 3 }}{\text{ and }}x = \frac{1}{{\sqrt 3 }} \cr & {\text{local minimum at }}x = 0 \cr & {\text{local maximum at }}x = - \frac{1}{{\sqrt 3 }}{\text{ and }}x = \frac{1}{{\sqrt 3 }} \cr} $$