Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 24

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts: }}\left( {0,0} \right),\left( {27,0} \right) \cr & {\text{Local maximum at }}\left( {0,0} \right) \cr & {\text{Local minimum at }}\left( {8, - 4} \right) \cr & {\text{No inflection points}} \cr & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = x - 3{x^{2/3}} \cr & {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \left( 0 \right) - 3{\left( 0 \right)^{2/3}} \cr & f\left( 0 \right) = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = x - 3{x^{2/3}} \cr & x - 3{x^{2/3}} = 0 \cr & x\left( {1 - 3{x^{ - 1/3}}} \right) = 0 \cr & x = 0,{\text{ }}\frac{3}{{{x^{1/3}}}} = 1 \to x = 27 \cr & x{\text{ - intercepts: }}\left( {0,0} \right),\left( {27,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x - 3{x^{2/3}}} \right] \cr & f'\left( x \right) = 1 - 2{x^{ - 1/3}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & 1 - 2{x^{ - 1/3}} = 0 \cr & {x^{ - 1/3}}\left( {{x^{1/3}} - 2} \right) = 0 \cr & {x^{ - 1/3}} = 0,{\text{ }}{x^{1/3}} - 2 = 0 \cr & {\text{The derivate is undefined at }}x = 0 \cr & {\text{ }}{x^{1/3}} - 2 = 0 \cr & x = {\left( 2 \right)^3} = 8 \cr & {\text{The critical points are }}x = 0{\text{ and }}x = 8. \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {1 - 2{x^{ - 1/3}}} \right] \cr & f''\left( x \right) = \frac{2}{3}{x^{ - 4/3}} \cr & {\text{Evaluate }}f''\left( 0 \right){\text{ at the critical point }}x = 8 \cr & f''\left( 0 \right) = \frac{2}{3}{\left( 0 \right)^{ - 4/3}} = 0,{\text{ The second derivative test is inconclusive}} \cr & {\text{using the first derivative test:}} \cr & f'\left( { - 1} \right) = 1 - 2{\left( { - 1} \right)^{ - 1/3}} = 3,{\text{ increasing}} \cr & f'\left( 0 \right) = 1 - 2{\left( 0 \right)^{ - 1/3}}{\text{, undefined but }}x = 0\,{\text{is in the domain of }}f \cr & f'\left( 2 \right) = 1 - 2{\left( 2 \right)^{ - 1/3}} \approx - 0.587,{\text{ decreasing}} \cr & {\text{By the fist derivative test we can conclude that there is a local}} \cr & {\text{maximum at }}\left( {0,f\left( 0 \right)} \right) = \left( {0,0} \right) \cr & and \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical point }}x = 8 \cr & f''\left( 8 \right) = \frac{2}{3}{\left( { - 8} \right)^{ - 4/3}} = \frac{1}{{24}} > 0 \to {\text{local minimum at }}\left( {8,f\left( 8 \right)} \right) \cr & \left( {8,f\left( 8 \right)} \right) \to \left( {8, - 4} \right) \cr & {\text{Local minimum at }}\left( {8, - 4} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{2}{3}{x^{ - 4/3}} = 0 \cr & {\text{The second derivative is not defined at }}x = 0 \cr & f''\left( { - 1} \right) = \frac{2}{3} > 0,{\text{ }}\left( {{\text{Concave up}}} \right) \cr & f''\left( 1 \right) = \frac{2}{3} > 0,{\text{ }}\left( {{\text{Concave up}}} \right) \cr & {\text{Then}}{\text{, no inflection points}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {x - 3{x^{2/3}}} \right) = + \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {x - 3{x^{2/3}}} \right) = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$