Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 75

$$\eqalign{ & x{\text{ - intercept: none}} \cr & y{\text{ - intercept: none}} \cr & {\text{Relative minimum at }}\left( {1,6} \right) \cr & {\text{Relative maximum at }}\left( { - 1, - 6} \right) \cr & {\text{Inflection points: none}} \cr & {\text{Vertical asymptote: }}x = 0 \cr & {\text{Horizontal asymptote: none}} \cr} $$

$$\eqalign{ & f\left( x \right) = {x^3} + x + \frac{4}{x} \cr & {\text{*Find the }}y{\text{ intercept, let }}x = 0 \cr & f\left( x \right) = {0^3} + 0 + \frac{4}{0} \cr & {\text{There are no }}y{\text{ - intercepts}} \cr & {\text{*Find the }}x{\text{ intercept, let }}y = 0 \cr & {x^3} + x + \frac{4}{x} = 0 \cr & {x^4} + {x^2} + 4 = 0 \cr & {\text{No real solutions, then there are no }}x{\text{ intercepts}}{\text{.}} \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {{x^3} + x + \frac{4}{x}} \right] \cr & y' = 3{x^2} + 1 - \frac{4}{{{x^2}}} \cr & {\text{Let }}y' = 0 \cr & 3{x^2} + 1 - \frac{4}{{{x^2}}} = 0 \cr & 3{x^4} + {x^2} - 4 = 0 \cr & \left( {3{x^2} + 4} \right)\left( {{x^2} - 1} \right) = 0 \cr & 3{x^2} + 4 = 0 \to {\text{ No real solutions}}{\text{.}} \cr & {x^2} - 1 = 0 \to x = - 1,{\text{ }}x = 1 \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {3{x^2} + 1 - \frac{4}{{{x^2}}}} \right] \cr & y'' = 6x + \frac{8}{{{x^3}}} \cr & \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = - 1 \cr & y''\left( { - 1} \right) = - 14 < 0,{\text{ there is a relative maximum at }}f\left( { - 1} \right) \cr & {\text{ }}f\left( { - 1} \right) = - 6 \cr & {\text{Relative maximum at }}\left( { - 1, - 6} \right) \cr & \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = 1 \cr & y''\left( 1 \right) = 14 > 0,{\text{ there is a relative minimum at }}f\left( 1 \right) \cr & {\text{ }}f\left( 1 \right) = 6 \cr & {\text{Relative minimum at }}\left( {1,6} \right) \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & 6x + \frac{8}{{{x^3}}} = 0 \cr & 6{x^4} + 8 = 0,{\text{ No real solutions, then }} \cr & {\text{There are no inflection points}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {x^3} + x + \frac{4}{x} \cr & x = 0 \cr & {\text{Vertical asymptotes at }}x = 0 \cr & \mathop {\lim }\limits_{x \to \infty } \left( {{x^3} + x + \frac{4}{x}} \right) = \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + x + \frac{4}{x}} \right) = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr } $$