Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 19

$$\eqalign{ & y{\text{ - intercept }}\left( {0,12} \right) \cr & {\text{no }}x{\text{ - intercepts}} \cr & {\text{Relative minimum at }}\left( {3,3} \right) \cr & {\text{Relative maximum at }}\left( { - 4, - 4} \right) \cr & {\text{No inflection points}} \cr & {\text{Vertical asymptote }}x = - \frac{1}{2} \cr & {\text{Slant asymptote }}\frac{x}{2} - \frac{1}{4} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^2} + 12}}{{2x + 1}} \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \frac{{{{\left( 0 \right)}^2} + 12}}{{2\left( 0 \right) + 1}} \cr & f\left( 0 \right) = 12 \cr & y{\text{ - intercept }}\left( {0,12} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = \frac{{{x^2} + 12}}{{2x + 1}} \cr & {x^2} + 12 = 0,{\text{ No real solutions}}{\text{, then}} \cr & {\text{There are no }}x{\text{ - intercepts}} \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2} + 12}}{{2x + 1}}} \right] \cr & f'\left( x \right) = \frac{{\left( {2x + 1} \right)\left( {2x} \right) - \left( {{x^2} + 12} \right)\left( 2 \right)}}{{{{\left( {2x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{4{x^2} + 2x - 2{x^2} - 24}}{{{{\left( {2x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^2} + 2x - 24}}{{{{\left( {2x + 1} \right)}^2}}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & \frac{{2{x^2} + 2x - 24}}{{{{\left( {2x + 1} \right)}^2}}} = 0 \cr & 2{x^2} + 2x - 24 = 0 \cr & {x^2} + x - 12 = 0 \cr & \left( {x + 4} \right)\left( {x - 3} \right) = 0 \cr & x = - 4,{\text{ }}x = 3 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2{x^2} + 2x - 24}}{{{{\left( {2x + 1} \right)}^2}}}} \right] \cr & f''\left( x \right) = \frac{{{{\left( {2x + 1} \right)}^2}\left( {4x + 2} \right) - 2\left( {2{x^2} + 2x - 24} \right)\left( {2x + 1} \right)\left( 2 \right)}}{{{{\left( {2x + 1} \right)}^4}}} \cr & {\text{Simplifying}} \cr & f''\left( x \right) = \frac{{\left( {2x + 1} \right)\left( {4x + 2} \right) - 2\left( {2{x^2} + 2x - 24} \right)\left( 2 \right)}}{{{{\left( {2x + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{8{x^2} + 4x + 4x + 2 - 8{x^2} - 8x + 96}}{{{{\left( {2x + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{98}}{{{{\left( {2x + 1} \right)}^3}}} \cr & \cr & {\text{Evaluate the second derivative at the critical points}} \cr & *f''\left( { - 4} \right) = \frac{{98}}{{{{\left( {2\left( { - 4} \right) + 1} \right)}^3}}} = - \frac{2}{7} < 0 \cr & {\text{Relative maximum at }}\left( { - 4,f\left( { - 4} \right)} \right) \to \left( { - 4, - 4} \right) \cr & *f''\left( 3 \right) = \frac{{98}}{{{{\left( {2\left( 3 \right) + 1} \right)}^3}}} = \frac{2}{7} > 0 \cr & {\text{Relative minimum at }}\left( {3,f\left( 3 \right)} \right) \to \left( {3,3} \right) \cr & \cr & *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{{98}}{{{{\left( {2x + 1} \right)}^3}}} = 0 \cr & {\text{There are no values at which }}f''\left( x \right) = 0 \cr & {\text{No inflection points}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & f\left( x \right) = \frac{{{x^2} + 12}}{{2x + 1}} \cr & 2x + 1 = 0 \to x = - \frac{1}{2} \cr & {\text{Vertical asymptote }}x = - \frac{1}{2} \cr & \cr & *{\text{Use long division}} \cr & f\left( x \right) = \frac{{{x^2} + 12}}{{2x + 1}} = \frac{x}{2} - \frac{1}{4} + \frac{{49/4}}{{2x + 1}} \cr & {\text{Slant asymptote }}\frac{x}{2} - \frac{1}{4} \cr & \cr & {\text{Graph}} \cr} $$