Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 16

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - Intercepts }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & {\text{Relative maximum at }}\left( {0,0} \right) \cr & {\text{No inflection points}} \cr & {\text{Vertical asymptotes }}x = 2,{\text{ }}x = - 2 \cr & {\text{Horizontal asymptote }}y = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{{x^2} - 4}} \cr & {\text{Find the }}y{\text{ intercept}}{\text{, }}x = 0 \cr & f\left( 0 \right) = \frac{{{0^2}}}{{{0^2} - 4}} = 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{{{x^2}}}{{{x^2} - 4}} \cr & {x^2} - 4 = 0 \cr & x = \pm 2 \cr & x{\text{ - Intercepts }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{x^2} - 4}}} \right] \cr & f'\left( x \right) = \frac{{\left( {{x^2} - 4} \right)\left( {2x} \right) - \left( {{x^2}} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^3} - 8x - 2{x^3}}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - 8x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & {\text{Let }}f'\left( x \right) = 0 \cr & \frac{{ - 8x}}{{{{\left( {{x^2} - 4} \right)}^2}}} = 0 \cr & x = 0 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 8x}}{{{{\left( {{x^2} - 4} \right)}^2}}}} \right] \cr & f''\left( x \right) = \frac{{{{\left( {{x^2} - 4} \right)}^2}\left( { - 8} \right) + 8x\left( 2 \right)\left( {{x^2} - 4} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^4}}} \cr & f''\left( x \right) = \frac{{\left( {{x^2} - 4} \right)\left( { - 8} \right) + 8x\left( 2 \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr & f''\left( x \right) = \frac{{ - 8{x^2} + 32 - 32{x^2}}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr & f''\left( x \right) = \frac{{24{x^2} + 32}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical point }}x = 0 \cr & f''\left( 0 \right) = \frac{{24{{\left( 0 \right)}^2} + 32}}{{{{\left( {{{\left( 0 \right)}^2} - 4} \right)}^3}}} = - \frac{1}{2} < 0 \cr & {\text{Relative maximum at }}\left( {0,f\left( 0 \right)} \right) \to \left( {0,0} \right) \cr & \cr & *{\text{Find the Inflection points let }}f''\left( x \right) = 0 \cr & \frac{{24{x^2} + 32}}{{{{\left( {{x^2} - 4} \right)}^3}}} = 0 \cr & \frac{{30{x^2} + 40}}{{{{\left( {{x^2} - 4} \right)}^3}}} = 0 \cr & 24{x^2} + 32 = 0,{\text{ has no real solutions}}{\text{, then}} \cr & {\text{No inflection points}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & y = \frac{{{x^2}}}{{{x^2} - 4}} \cr & {x^2} - 4 = 0,{\text{ }}x = \pm 2 \cr & {\text{Vertical asymptotes}}{\text{, }}x = 2,{\text{ }}x = - 2 \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2}}}{{{x^2} - 4}}} \right) = 1 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2}}}{{{x^2} - 4}}} \right) = 1 \cr & {\text{Horizontal asymptote }}y = 1 \cr & \cr & {\text{Graph}} \cr} $$