Chapter 3 - Applications of Differentiation - 3.6 Exercises - Page 212: 12

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Relative maximum at }}\left( { - 3\sqrt 3 , - \frac{{9\sqrt 3 }}{2}} \right) \cr & {\text{Relative minimum at }}\left( {3\sqrt 3 ,\frac{{9\sqrt 3 }}{2}} \right) \cr & {\text{Inflection point at }}\left( {0,0} \right) \cr & {\text{Vertical asymptotes at }}x = \pm 3 \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^3}}}{{{x^2} - 9}} \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & y = \frac{{{0^3}}}{{{0^2} - 9}} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}y = 0 \cr & 0 = \frac{{{x^3}}}{{{x^2} - 9}} \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {\frac{{{x^3}}}{{{x^2} - 9}}} \right] \cr & y' = \frac{{\left( {{x^2} - 9} \right)\left( {3{x^2}} \right) - {x^3}\left( {2x} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}} \cr & y' = \frac{{3{x^4} - 27{x^2} - 2{x^4}}}{{{{\left( {{x^2} - 9} \right)}^2}}} \cr & y' = \frac{{{x^4} - 27{x^2}}}{{{{\left( {{x^2} - 9} \right)}^2}}} \cr & {\text{Let }}y' = 0 \cr & {x^4} - 27{x^2} = 0 \cr & {x^2}\left( {{x^2} - 27} \right) = 0 \cr & x = 0,{\text{ }}x = \pm 3\sqrt 3 \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {\frac{{{x^4} - 27{x^2}}}{{{{\left( {{x^2} - 9} \right)}^2}}}} \right] \cr & {\text{Using a CAS }} \cr & y'' = \frac{{18x\left( {{x^2} + 27} \right)}}{{{{\left( {{x^2} - 9} \right)}^3}}} \cr & \cr & {\text{Evaluate }}y''\left( 0 \right) \cr & *y''\left( 0 \right) = 0,{\text{ inconclusive test}}{\text{, use the first derivative test}} \cr & y'\left( { - 1} \right) < 0,{\text{ }}y'\left( 1 \right) < 0,{\text{ No relative extrema at }}x = 0 \cr & *y''\left( { - 3\sqrt 3 } \right) = \frac{{18\left( { - 3\sqrt 3 } \right)\left( {{{\left( { - 3\sqrt 3 } \right)}^2} + 27} \right)}}{{{{\left( {{{\left( { - 3\sqrt 3 } \right)}^2} - 9} \right)}^3}}} < 0,{\text{ then}} \cr & {\text{Relative maximum at }}\left( { - 3\sqrt 3 ,f\left( { - 3\sqrt 3 } \right)} \right) \to \left( { - 3\sqrt 3 , - \frac{{9\sqrt 3 }}{2}} \right) \cr & y''\left( {3\sqrt 3 } \right) = \frac{{18\left( { - 3\sqrt 3 } \right)\left( {{{\left( { - 3\sqrt 3 } \right)}^2} + 27} \right)}}{{{{\left( {{{\left( { - 3\sqrt 3 } \right)}^2} - 9} \right)}^3}}} > 0,{\text{ then}} \cr & {\text{Relative minimum at }}\left( {3\sqrt 3 ,f\left( {3\sqrt 3 } \right)} \right) \to \left( {3\sqrt 3 ,\frac{{9\sqrt 3 }}{2}} \right) \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & \frac{{18x\left( {{x^2} + 27} \right)}}{{{{\left( {{x^2} - 9} \right)}^3}}} = 0 \cr & x = 0 \cr & {\text{Inflection point at }}\left( {0,0} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & \frac{{{x^3}}}{{{x^2} - 9}} \cr & {x^2} - 9 = 0 \cr & x = - 3,{\text{ }}x = 3 \cr & {\text{Vertical asymptotes at }}x = \pm 3 \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^3}}}{{{x^2} - 9}}} \right) = \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^3}}}{{{x^2} - 9}}} \right) = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$