Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 82

$$\eqalign{ & {\text{Critical points: }}x = 0{\text{ and }}x = \frac{5}{3} \cr & {\text{local minimum at }}x = \frac{5}{3} \cr & {\text{Neither a local minimum nor a local maximum at }}x = 0 \cr} $$

$$\eqalign{ & f\left( x \right) = \sqrt x \left( {\frac{{12}}{7}{x^3} - 4{x^2}} \right) \cr & f\left( x \right) = \frac{{12}}{7}{x^{7/2}} - 4{x^{5/2}} \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{12}}{7}{x^{7/2}} - 4{x^{5/2}}} \right] \cr & f'\left( x \right) = 6{x^{5/2}} - 10{x^{3/2}} \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & 6{x^{5/2}} - 10{x^{3/2}} = 0 \cr & 2{x^{3/2}}\left( {3x - 5} \right) = 0 \cr & {\text{The critical points are }}x = 0{\text{ and }}x = \frac{5}{3} \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {6{x^{5/2}} - 10{x^{3/2}}} \right] \cr & f''\left( x \right) = 15{x^{3/2}} - 15{x^{1/2}} \cr & \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}f''\left( 0 \right) \cr & f''\left( 0 \right) = 15{\left( 0 \right)^{3/2}} - 15{\left( 0 \right)^{1/2}} \cr & f''\left( 0 \right) = 0 \cr & f''\left( 0 \right) = 0,{\text{ then the test is inconclusive; may have a local maximum}}{\text{,}} \cr & {\text{local minimum}}{\text{, or neither at 0}}{\text{.}} \cr & {\text{Evaluating }}f'\left( { - 1} \right){\text{ and }}f'\left( 1 \right) \cr & {\text{We cannot evaluate at }}x = - 1,{\text{ because is not in the domain of }}f\left( x \right). \cr & {\text{So}}{\text{, the first derivative does not change sign at }}x = 0,{\text{ which means }}f \cr & {\text{does not have a local maximum or minimum at }}x = 0 \cr & \cr & {\text{Evaluate }}f''\left( {\frac{5}{3}} \right) \cr & f''\left( {\frac{5}{3}} \right) = 15{\left( {\frac{5}{3}} \right)^{3/2}} - 15{\left( {\frac{5}{3}} \right)^{1/2}} \cr & f''\left( {\frac{5}{3}} \right) \approx 12.9 \cr & f''\left( {\frac{5}{3}} \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local minimun at }}x = \frac{5}{3} \cr & \cr & {\text{Critical points: }}x = 0{\text{ and }}x = \frac{5}{3} \cr & {\text{local minimum at }}x = \frac{5}{3} \cr & {\text{Neither a local minimum nor a local maximum at }}x = 0 \cr} $$