Chapter 3 - Applications of Differentiation - 3.5 Exercises - Page 203: 76

$$\eqalign{ & D:R \ne \left\{ { - 1,2} \right\} \cr & {\text{Range: }}\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right) \cr & {\text{Vertical asymptotes: }}x = - 1{\text{ and }}x = 2 \cr & {\text{Horizontal asymptote: }}y = 0 \cr & {\text{Relative maximum: }}\left( {\frac{1}{2}, - \frac{4}{9}} \right) \cr & {\text{No Inflection points}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{1}{{{x^2} - x - 2}} \cr & {\text{From the graph }}\left( {{\text{Using Geogebra}}} \right): \cr & {\text{Domain: all real numbers except }}x = - 1{\text{ and }}x = 2 \cr & D:R \ne \left\{ { - 1,2} \right\} \cr & {\text{Range: }}\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right) \cr & {\text{Vertical asymptotes: }}x = - 1{\text{ and }}x = 2 \cr & {\text{Horizontal asymptote: }}y = 0 \cr & \cr & {\text{Calculate derivatives}} \cr & f'\left( x \right) = - {\left( {{x^2} - x - 2} \right)^{ - 2}}\left( {2x - 1} \right) \cr & f'\left( x \right) = 0 \to 2x - 1 = 0,{\text{ }}x = \frac{1}{2} \cr & {\text{Relative maximum at }}\left( {\frac{1}{2},f\left( {\frac{1}{2}} \right)} \right):\left( {\frac{1}{2}, - \frac{4}{9}} \right) \cr & f''\left( x \right) = \frac{{6\left( {{x^2} - x + 1} \right)}}{{{{\left( {{x^2} - x - 2} \right)}^3}}} \cr & f''\left( x \right) = 0 \to {x^2} - x + 1 = 0,{\text{ No real solutions}} \cr & {\text{No Inflection points}} \cr & \cr & {\text{There are no values at which }}f'\left( x \right) = 0,{\text{ no relative extrema}} \cr & f''\left( x \right) = 10\left( { - 3{x^{ - 4}}} \right) = - \frac{{30}}{{{x^4}}},{\text{ }} \cr & {\text{There are no values at which }}f''\left( x \right) = 0,{\text{ no inflection points}} \cr & \cr & {\text{Graph}} \cr} $$