Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 17

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{No relative extrema}}{\text{.}} \cr & {\text{Inflection point}}\left( {0,0} \right) \cr & {\text{Vertical asymptotes }}x = 1,{\text{ }}x = - 1 \cr & {\text{Horizontal asymptote }}y = 0 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{3x}}{{{x^2} - 1}} \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} - 1}} \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} - 1}} \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} - 1}}} \right] \cr & f'\left( x \right) = \frac{{\left( {{x^2} - 1} \right)\left( 3 \right) - 3x\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{3{x^2} - 3 - 6{x^2}}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - 3{x^2} - 3}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & {\text{Let }}f'\left( x \right) = 0 \cr & - 3{x^2} - 3 = 0 \cr & 3{x^2} + 3 = 0,{\text{ there are no values at which }}f'\left( x \right) = 0, \cr & {\text{No relative extrema}}{\text{.}} \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 3{x^2} - 3}}{{{{\left( {{x^2} - 1} \right)}^2}}}} \right] \cr & f''\left( x \right) = \frac{{{{\left( {{x^2} - 1} \right)}^2}\left( { - 6x} \right) - \left( { - 3{x^2} - 3} \right)\left( 2 \right)\left( {{x^2} - 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^4}}} \cr & f''\left( x \right) = \frac{{6x\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 1} \right)}^4}}} \cr & \cr & *{\text{Find the Inflection points}} \cr & f''\left( x \right) = \frac{{6x\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 1} \right)}^4}}} \cr & \frac{{6x\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 1} \right)}^4}}} = 0 \cr & x = 0 \cr & {\text{Inflection point at }}x = 0 \cr & f\left( 0 \right) = \frac{{3\left( 0 \right)}}{{{{\left( 0 \right)}^2} - 1}} = 0,{\text{ }}\left( {0,0} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & f\left( x \right) = \frac{{3x}}{{{x^2} - 1}} \cr & {x^2} - 1 = 0,{\text{ }}x = \pm 1 \cr & {\text{Vertical asymptotes}}{\text{, }}x = 1,{\text{ }}x = - 1 \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{3x}}{{{x^2} - 1}}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{3x}}{{{x^2} - 1}}} \right) = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr & {\text{Graph}} \cr} $$