Chapter 3 - Applications of Differentiation - 3.5 Exercises - Page 203: 77

$$\eqalign{ & {\text{Inflection point }}\left( {2,0} \right) \cr & {\text{Vertical asymptotes: }}x = 1,{\text{ }}x = 3 \cr & {\text{Horizontal asymptote at }}y = 0 \cr & {\text{No relative extrema}} \cr & {\text{Domain}}:\left( { - \infty ,1} \right),\left( {1,3} \right),\left( {3,\infty } \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{x - 2}}{{{x^2} - 4x + 3}} \cr & {x^2} - 4x + 3 \ne 0 \cr & \left( {x - 3} \right)\left( {x - 1} \right) \ne 0 \cr & x \ne 1,{\text{ }}x \ne 3 \cr & {\text{Then, the domain of the function is:}} \cr & D:\left( { - \infty ,1} \right),\left( {1,3} \right),\left( {3,\infty } \right) \cr & \cr & {\text{*Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{x - 2}}{{{x^2} - 4x + 3}}} \right] \cr & f'\left( x \right) = \frac{{\left( {{x^2} - 4x + 3} \right)\left( 1 \right) - \left( {x - 2} \right)\left( {2x - 4} \right)}}{{{{\left( {{x^2} - 4x + 3} \right)}^2}}} \cr & f'\left( x \right) = \frac{{{x^2} - 4x + 3 - 2{x^2} + 4x + 4x - 8}}{{{{\left( {{x^2} - 4x + 3} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - {x^2} + 4x - 5}}{{{{\left( {{x^2} - 4x + 3} \right)}^2}}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & \frac{{ - {x^2} + 4x - 5}}{{{{\left( {{x^2} - 4x + 3} \right)}^2}}} = 0 \cr & - {x^2} + 4x - 5 = 0 \cr & {\text{The quadratic equation has ho real solutions}}{\text{.}} \cr & {\text{Then there are no relative extrema}}{\text{.}} \cr & \cr & {\text{*The vertical asymptotes are at }}{x^2} - 4x + 3 = 0,{\text{ we obtain}} \cr & {\text{vertical asymptotes at: }}x = 1,{\text{ }}x = 3 \cr & \cr & *{\text{Find the horizontal asymptotes}} \cr & \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{x - 2}}{{{x^2} - 4x + 3}} = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{x - 2}}{{{x^2} - 4x + 3}} = 0 \cr & {\text{Horizontal asymptote at }}y = 0 \cr & \cr & {\text{*Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - {x^2} + 4x - 5}}{{{{\left( {{x^2} - 4x + 3} \right)}^2}}}} \right] \cr & {\text{Using a CAS we obtain}} \cr & f''\left( x \right) = \frac{{2\left( {x - 2} \right)\left( {{x^2} - 4x + 7} \right)}}{{{{\left( {{x^2} - 4x + 3} \right)}^3}}} \cr & {\text{Set }}f''\left( x \right) = 0 \cr & 2\left( {x - 2} \right)\underbrace {\left( {{x^2} - 4x + 7} \right)}_{{\text{No real solutions}}} = 0 \cr & x = 2 \cr & {\text{For }}\left( {1,2} \right) \to f''\left( {\frac{3}{2}} \right) = \frac{{16}}{{27}} > 0 \cr & {\text{For }}\left( {2,3} \right) \to f''\left( {\frac{5}{2}} \right) = - \frac{{16}}{{27}} < 0 \cr & {\text{Then there is an inflection point at }}x = 2 \cr & f\left( 2 \right) = 0 \to {\text{Inflection point }}\left( {2,0} \right) \cr & \cr \cr} $$