Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 76

$$\eqalign{ & x{\text{ - intercept: }}\left( { - 1,0} \right) \cr & y{\text{ - intercept: none}} \cr & {\text{Relative minimum at }}\left( {\frac{1}{{\root 3 \of 2 }},\root 3 \of 4 + \frac{1}{{\root 3 \of 2 }}} \right) \cr & {\text{Inflection points: }}\left( { - 1,0} \right) \cr & {\text{Vertical asymptote: }}x = 0 \cr & {\text{Horizontal asymptote: none}} \cr} $$

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