Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 63

$$\eqalign{ & \left. a \right)x = 8 \cr & \left. b \right) \cr & {\text{Abs}}{\text{. max 3}}\sqrt 2 {\text{ at }}x = 6{\text{ and }}x = 12 \cr & {\text{Abs}}{\text{. min }}4{\text{ at }}x = 8 \cr & \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{x}{{\sqrt {x - 4} }}{\text{ on }}\left[ {6,12} \right] \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{\sqrt {x - 4} }}} \right] \cr & f'\left( x \right) = \frac{{\sqrt {x - 4} - x\left( {\frac{1}{{2\sqrt {x - 4} }}} \right)}}{{{{\left( {\sqrt {x - 4} } \right)}^2}}} \cr & f'\left( x \right) = \frac{{\frac{{2x - 8 - x}}{{2\sqrt {x - 4} }}}}{{x - 4}} \cr & f'\left( x \right) = \frac{{x - 8}}{{2{{\left( {x - 4} \right)}^{3/2}}}} \cr & \cr & a.{\text{ Find the critical points}}{\text{, let }}f'\left( x \right) = 0 \cr & \frac{{x - 8}}{{2{{\left( {x - 4} \right)}^{3/2}}}} \cr & {\text{The derivative is undefined at }}x = 4,\,{\text{but }}4\,{\text{is not in}} \cr & {\text{the interval }}\left[ {6,12} \right] \cr & x - 8 = 0 \cr & x = 8 \cr & \cr & {\text{The critical point is }}x = 8 \cr & \cr & \left( {\text{b}} \right){\text{Evaluate }}f\left( x \right){\text{ at the critical points and the endpoints }} \cr & f\left( 6 \right) = \frac{6}{{\sqrt {6 - 4} }} = 3\sqrt 2 \approx 4.24,{\text{ }}\left( {{\text{Largest}}} \right) \cr & f\left( 8 \right) = \frac{8}{{\sqrt {8 - 4} }} = 4,{\text{ }}\left( {{\text{Smallest}}} \right) \cr & f\left( {12} \right) = \frac{{12}}{{\sqrt {12 - 4} }} = 3\sqrt 2 \approx 4.24,{\text{ }}\left( {{\text{Largest}}} \right) \cr & \cr & {\text{Therefore}}{\text{,}} \cr & {\text{*The absolute maximum of }}f\left( x \right){\text{ is 3}}\sqrt 2 {\text{ at }}x = 6{\text{ and }}x = 12 \cr & {\text{*The absolute minimum of }}f\left( x \right){\text{ is 4 at }}x = 8 \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$