Chapter 4 - Applications of the Derivative - Review Exercises - Page 330: 14

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Relative minimum at }}\left( { - \sqrt 3 , - \frac{{\sqrt 3 }}{2}} \right) \cr & {\text{Relative maximum at }}\left( {\sqrt 3 ,\frac{{\sqrt 3 }}{2}} \right) \cr & {\text{inflection points }}\left( { - 3, - \frac{3}{4}} \right),\left( {0,0} \right),\left( {3,\frac{3}{4}} \right) \cr & {\text{No vertical asymptotes}} \cr & {\text{Horizontal asymptote }}y = 0 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{3x}}{{{x^2} + 3}} \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \frac{{3\left( 0 \right)}}{{{{\left( 0 \right)}^2} + 3}} \cr & f\left( 0 \right) = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = \frac{{3x}}{{{x^2} + 3}} \cr & 3x = 0 \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{3x}}{{{x^2} + 3}}} \right] \cr & f'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( 3 \right) - 3x\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr & f'\left( x \right) = \frac{{3{x^2} + 9 - 6{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr & f'\left( x \right) = \frac{{9 - 3{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & \frac{{9 - 3{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} = 0 \cr & 9 - 3{x^2} = 0 \cr & {x^2} = 3 \cr & x = - \sqrt 3 ,{\text{ }}x = \sqrt 3 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{9 - 3{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}}} \right] \cr & f''\left( x \right) = \frac{{{{\left( {{x^2} + 3} \right)}^2}\left( { - 6x} \right) - 4x\left( {9 - 3{x^2}} \right)\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} \cr & {\text{Simplifying}} \cr & f''\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( { - 6x} \right) - 4x\left( {9 - 3{x^2}} \right)}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr & f''\left( x \right) = \frac{{ - 6{x^3} - 18x - 36x + 12{x^3}}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr & f''\left( x \right) = \frac{{6{x^3} - 54x}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr & \cr & {\text{Evaluate the second derivative at the critical points}} \cr & *f''\left( { - \sqrt 3 } \right) = \frac{{6{{\left( { - \sqrt 3 } \right)}^3} - 54\left( { - \sqrt 3 } \right)}}{{{{\left( {{{\left( { - \sqrt 3 } \right)}^2} + 3} \right)}^3}}} = \frac{{\sqrt 3 }}{6} > 0 \cr & f\left( { - \sqrt 3 } \right) = \frac{{3\left( { - \sqrt 3 } \right)}}{{{{\left( { - \sqrt 3 } \right)}^2} + 3}} = - \frac{{\sqrt 3 }}{2} \cr & {\text{Relative minimum at }}\left( { - \sqrt 3 ,f\left( { - \sqrt 3 } \right)} \right) \to \left( { - \sqrt 3 , - \frac{{\sqrt 3 }}{2}} \right) \cr & \cr & *f''\left( {\sqrt 3 } \right) = \frac{{6{{\left( {\sqrt 3 } \right)}^3} - 54\left( {\sqrt 3 } \right)}}{{{{\left( {{{\left( {\sqrt 3 } \right)}^2} + 3} \right)}^3}}} = - \frac{{\sqrt 3 }}{6} < 0 \cr & f\left( {\sqrt 3 } \right) = \frac{{3\left( {\sqrt 3 } \right)}}{{{{\left( {\sqrt 3 } \right)}^2} + 3}} = \frac{{\sqrt 3 }}{2} \cr & {\text{Relative maximum at }}\left( {\sqrt 3 ,f\left( {\sqrt 3 } \right)} \right) \to \left( {\sqrt 3 ,\frac{{\sqrt 3 }}{2}} \right) \cr & \cr & *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{{6{x^3} - 54x}}{{{{\left( {{x^2} + 3} \right)}^3}}} = 0 \cr & 6{x^3} - 54x = 0 \cr & 6x\left( {{x^2} - 9} \right) = 0 \cr & x = 0,{\text{ }}x = - 3,{\text{ }}x = 3 \cr & f\left( { - 3} \right) = \frac{{3\left( { - 3} \right)}}{{{{\left( { - 3} \right)}^2} + 3}} = - \frac{3}{4},{\text{ }} \cr & f\left( 0 \right) = \frac{{3\left( 0 \right)}}{{{{\left( 0 \right)}^2} + 3}} = 0,{\text{ }} \cr & f\left( 3 \right) = \frac{{3\left( 3 \right)}}{{{{\left( 3 \right)}^2} + 3}} = \frac{3}{4} \cr & {\text{The inflection points are:}} \cr & \left( { - 3, - \frac{3}{4}} \right),\left( {0,0} \right),\left( {3,\frac{3}{4}} \right) \cr & {\text{*Calculate the asymptotes}} \cr & f\left( x \right) = \frac{{3x}}{{{x^2} + 3}} \cr & {\text{The denominator is never zero}}{\text{, then there are no}} \cr & {\text{vertical asymptotes}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{3x}}{{{x^2} + 3}}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{3x}}{{{x^2} + 3}}} \right) = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr & {\text{Graph}} \cr} $$