Chapter 3 - Applications of Differentiation - 3.5 Exercises - Page 203: 73

$$\eqalign{ & {\text{Domain:}}\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right) \cr & {\text{Intercepts: None}} \cr & {\text{Symmetry:Origin}} \cr & {\text{Vertical asymptotes at: }}x = \pm 2 \cr & {\text{Relative maximum at }}\left( { - \sqrt 6 , - 6\sqrt 3 } \right) \cr & {\text{Relative minimum at }}\left( {\sqrt 6 ,6\sqrt 3 } \right) \cr} $$

$$\eqalign{ & y = \frac{{{x^3}}}{{\sqrt {{x^2} - 4} }} \cr & {\text{The domain is given for }}{x^2} - 4 > 0,{\text{ Then}} \cr & \left( {x + 2} \right)\left( {x - 2} \right) > 0 \cr & {\text{Domain: }}\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right) \cr & \cr & {\text{Find the intercepts}} \cr & *{\text{For }}y = 0 \cr & 0 = \frac{{{x^3}}}{{\sqrt {{x^2} - 4} }},{\text{ No solution in the domain}} \cr & {\text{Then there are no }}x{\text{ intercept}}{\text{.}} \cr & *{\text{ }}x = 0,{\text{It is not in the domain}}{\text{, then there are no }}y{\text{ intercepts}} \cr & {\text{Intercepts: none}} \cr & \cr & {\text{Let }}f\left( x \right) = \frac{{{x^3}}}{{\sqrt {{x^2} - 4} }} \cr & {\text{Test symmetry}} \cr & f\left( { - x} \right) = \frac{{{{\left( { - x} \right)}^3}}}{{\sqrt {{{\left( { - x} \right)}^2} - 4} }} \cr & f\left( { - x} \right) = \frac{{ - {x^3}}}{{\sqrt {{x^2} - 4} }} \cr & f\left( { - x} \right) = - \frac{{{x^3}}}{{\sqrt {{x^2} - 4} }} \cr & f\left( { - x} \right) = - f\left( x \right),{\text{ The function is odd}}{\text{, then the graph is}} \cr & {\text{symmetric with respect to the origin}}{\text{.}} \cr & \cr & {\text{Find the relative extrema}} \cr & {\text{First differentiate }}f\left( x \right) = \frac{{{x^3}}}{{\sqrt {{x^2} - 4} }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^3}}}{{\sqrt {{x^2} - 4} }}} \right] \cr & {\text{Using a graphing utility we obtain}} \cr & f'\left( x \right) = \frac{{2{x^2}\left( {{x^2} - 6} \right)}}{{{{\left( {{x^2} - 4} \right)}^{3/2}}}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to calculate the critical points}} \cr & 2{x^2}\left( {{x^2} - 6} \right) = 0 \cr & x = 0{\text{ }}\left( {{\text{It is not in the domain}}} \right),{\text{ }} \cr & {x^2} - 6 = 0 \to x = - \sqrt 6 ,{\text{ }}x = \sqrt 6 \cr & \cr & {\text{Test Intervals }} \cr & \left( { - \infty , - \sqrt 6 } \right),\left( { - \sqrt 6 , - 2} \right),\left( {2,\sqrt 6 } \right),\left( {\sqrt 6 ,\infty } \right) \cr & f'\left( { - 3} \right) > 0{\text{ and }}f'\left( { - 2.3} \right) < 0 \cr & {\text{There is a relative maximum at }}x = - \sqrt 6 \cr & f\left( { - \sqrt 6 } \right) = \frac{{{{\left( { - \sqrt 6 } \right)}^3}}}{{\sqrt {{{\left( { - \sqrt 6 } \right)}^2} - 4} }} = - 6\sqrt 3 ,{\text{ }}\left( { - \sqrt 6 , - 6\sqrt 3 } \right) \cr & \cr & f'\left( {2.1} \right) < 0,{\text{ }}f'\left( 3 \right) > 0 \cr & {\text{There is a relative minimum at }}x = \sqrt 6 \cr & f\left( {\sqrt 6 } \right) = \frac{{{{\left( {\sqrt 6 } \right)}^3}}}{{\sqrt {{{\left( {\sqrt 6 } \right)}^2} - 4} }} = 6\sqrt 3 ,{\text{ }}\left( {\sqrt 6 ,6\sqrt 3 } \right) \cr & \cr & {\text{The denominator is zero at }}x = - 2,{\text{ }}x = 2,{\text{ then}} \cr & {\text{Vertical asymptotes at: }}x = \pm 2 \cr & \cr & {\text{Summary:}} \cr & {\text{Domain:}}\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right) \cr & {\text{Intercepts: None}} \cr & {\text{Symmetry:Origin}} \cr & {\text{Vertical asymptotes at: }}x = \pm 2 \cr & {\text{Relative maximum at }}\left( { - \sqrt 6 , - 6\sqrt 3 } \right) \cr & {\text{Relative minimum at }}\left( {\sqrt 6 ,6\sqrt 3 } \right) \cr & \cr & {\text{Graph}} \cr} $$