Chapter 3 - The Derivative In Graphing And Applications - 3.3 Analysis Of Functions III: Rational Functions, Cusps, And Vertical Tangents - Exercises Set 3.3 - Page 215: 32

$$\eqalign{ & {\text{critical points: }}\left( {0,\root 3 \of { - 4} } \right),{\text{ }}\left( { - 2,0} \right),{\text{ }}\left( {2,0} \right) \cr & {\text{inflection points: }}\left( { - 2,0} \right),{\text{ }}\left( {2,0} \right) \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \root 3 \of {{x^2} - 4} \cr & {\text{Differentiate }}f'\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {{x^2} - 4} \right)}^{1/3}}} \right] \cr & f'\left( x \right) = \frac{1}{3}{\left( {{x^2} - 4} \right)^{ - 2/3}}\left( {2x} \right) \cr & f'\left( x \right) = \frac{{2x}}{{3{{\left( {{x^2} - 4} \right)}^{2/3}}}} \cr & {\text{Calculate the critical points set the derivative equal to 0}} \cr & \frac{{2x}}{{3{{\left( {{x^2} - 4} \right)}^{2/3}}}} = 0 \cr & 2x = 0 \cr & x = 0 \cr & {\text{The derivative is not defined when }}{x^2} - 4 = 0,{\text{ then}} \cr & {x^2} - 4 = 0 \cr & {x^2} = 4 \cr & x = \pm 2 \cr & {\text{Calculate }}f\left( 0 \right),{\text{ }}f\left( { \pm 2} \right) \cr & f\left( 0 \right) = \root 3 \of { - 4} ,{\text{ }}f\left( { - 2} \right) = 0,{\text{ }}f\left( 2 \right) = 0 \cr & {\text{Therefore, the critical points are:}} \cr & \left( {0,\root 3 \of { - 4} } \right),{\text{ }}\left( { - 2,0} \right),{\text{ }}\left( {2,0} \right) \cr & \cr & {\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x}}{{3{{\left( {{x^2} - 4} \right)}^{2/3}}}}} \right] \cr & f''\left( x \right) = \frac{{3{{\left( {{x^2} - 4} \right)}^{2/3}}\left( 2 \right) - 2x\left( 3 \right)\left( {\frac{2}{3}} \right){{\left( {{x^2} - 4} \right)}^{ - 1/3}}\left( {2x} \right)}}{{{{\left( {3{{\left( {{x^2} - 4} \right)}^{2/3}}} \right)}^2}}} \cr & f''\left( x \right) = \frac{{6{{\left( {{x^2} - 4} \right)}^{2/3}} - 8{x^2}{{\left( {{x^2} - 4} \right)}^{ - 1/3}}}}{{9{{\left( {{x^2} - 4} \right)}^{4/3}}}} \cr & f''\left( x \right) = \frac{{2{{\left( {{x^2} - 4} \right)}^{ - 1/3}}\left[ {3\left( {{x^2} - 4} \right) - 4{x^2}} \right]}}{{9{{\left( {{x^2} - 4} \right)}^{4/3}}}} \cr & f''\left( x \right) = \frac{{2\left( {3{x^2} - 12 - 4{x^2}} \right)}}{{9{{\left( {{x^2} - 4} \right)}^{ - 5/3}}}} \cr & f''\left( x \right) = - \frac{{2{{\left( {{x^2} - 4} \right)}^{5/3}}\left( {{x^2} + 12} \right)}}{9} \cr & {\text{Set }}f''\left( x \right) = 0 \cr & {x^2} - 4 = 0 \cr & x = \pm 2 \cr & {\text{Calculate }}f\left( { \pm 2} \right) \cr & f\left( 0 \right) = \root 3 \of { - 4} ,{\text{ }}f\left( { - 2} \right) = 0,{\text{ }}f\left( 2 \right) = 0 \cr & {\text{Therefore, the inflection points are:}} \cr & \left( { - 2,0} \right),{\text{ }}\left( {2,0} \right) \cr} $$