Chapter 3 - Applications of Differentiation - 3.6 Exercises - Page 212: 26

$$\eqalign{ & {\text{Domain:}}\left( { - \infty ,\infty } \right) \cr & {\text{Relative maximum at the point: }}(0.1292,{\text{4}}{\text{.063}}) \cr & {\text{Relative minimum at the point: }}(1.6085,{\text{2}}{\text{.7236}}) \cr & {\text{Vertical Asymptote: None}} \cr & {\text{Horizontal Asymptote: None}} \cr & {\text{Slant Asymptote: }}y = x \cr} $$

$$\eqalign{ & f\left( x \right) = x + \frac{4}{{{x^2} + 1}} \cr & {\text{The denominator is never zero}}{\text{, then the domain of the}} \cr & {\text{function is }}D:\left( { - \infty ,\infty } \right) \cr & \cr & {\text{*Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 0 + \frac{4}{{{{\left( 0 \right)}^2} + 1}} \cr & y = 4 \cr & y{\text{ - intercept }}\left( {0,4} \right) \cr & \cr & {\text{*Find the }}x{\text{ intercept}}{\text{, let }}y = 0 \cr & x + \frac{4}{{{x^2} + 1}} = 0 \cr & {x^3} + x + 4 = 0 \cr & {\text{Using technology we obtain the following real roots:}} \cr & x = - 1.3787 \cr & x{\text{ - intercepts }}\left( { - 1.3787,0} \right) \cr & \cr & *{\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x + \frac{4}{{{x^2} + 1}}} \right] \cr & f'\left( x \right) = 1 - \frac{{8x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{Let }}f'\left( x \right) = 0 \cr & 1 - \frac{{8x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0 \cr & \cr & {\text{Using technology we obtain the following real roots:}} \cr & {x_1} \approx 0.1292,{\text{ }}{x_2} = 1.6085 \cr & {\text{Graph using tecnology we obtain the graph shown below:}} \cr & {\text{Relative maximum at the point: }}(0.1292,{\text{4}}{\text{.063}}) \cr & {\text{Relative minimum at the point: }}(1.6085,{\text{2}}{\text{.7236}}) \cr & {\text{Vertical Asymptote: None}} \cr & {\text{Horizontal Asymptote: None}} \cr & \cr & {\text{Graph}} \cr} $$