Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 58

$$\eqalign{ & \left. a \right)x = 2 \cr & \left. b \right) \cr & {\text{Abs}}{\text{. max 0 at }}x = 0 \cr & {\text{Abs}}{\text{. min }} - 4.5254{\text{ at }}x = 2 \cr} $$

$$\eqalign{ & f\left( x \right) = {x^{1/2}}\left( {\frac{{{x^2}}}{5} - 4} \right){\text{ on }}\left[ {0,4} \right] \cr & f\left( x \right) = \frac{{{x^{5/2}}}}{5} - 4{x^{1/2}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^{5/2}}}}{5} - 4{x^{1/2}}} \right] \cr & f'\left( x \right) = \frac{{{x^{3/2}}}}{2} - 2{x^{ - 1/2}} \cr & f'\left( x \right) = \frac{1}{2}{x^{ - 1/2}}\left( {{x^2} - 4} \right) \cr & \cr & a.{\text{ Find the critical points}}{\text{, let }}f'\left( x \right) = 0 \cr & \frac{1}{2}{x^{ - 1/2}}\left( {{x^2} - 4} \right) = 0 \cr & {\text{The derivative is undefined at }}x = 0 \cr & \left( {x = 0\,{\text{is an endpoint}}} \right) \cr & {x^2} - 4 = 0 \cr & x = \pm 2 \cr & {\text{The interval defined for the function is }}\left[ {0,4} \right],{\text{ then}} \cr & {\text{The critical point is }}x = 2 \cr & \cr & \left( {\text{b}} \right){\text{Evaluate }}f\left( x \right){\text{ at the critical points and the endpoints }} \cr & f\left( 0 \right) = {\left( 0 \right)^{1/2}}\left( {\frac{{{{\left( 0 \right)}^2}}}{5} - 4} \right) = 0,{\text{ }}\left( {{\text{Largest}}} \right) \cr & f\left( 2 \right) = {\left( 2 \right)^{1/2}}\left( {\frac{{{{\left( 2 \right)}^2}}}{5} - 4} \right) \approx - 4.5254,{\text{ }}\left( {{\text{Smallest}}} \right) \cr & f\left( 4 \right) = {\left( 4 \right)^{1/2}}\left( {\frac{{{{\left( 4 \right)}^2}}}{5} - 4} \right) = - \frac{8}{5} = - 1.6 \cr & {\text{Therefore}}{\text{,}} \cr & {\text{*The absolute maximum of }}f\left( x \right){\text{ is 0 at }}x = 0 \cr & {\text{*The absolute minimum of }}f\left( x \right){\text{ is }} - 4.5254{\text{ at }}x = 2 \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$