Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 45

$$\eqalign{ & {\text{Relative maximum at }}\left( { - 3, - 12} \right) \cr & {\text{Relative minimum at }}\left( {3,12} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = 2x + \frac{{18}}{x} \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2x + \frac{{18}}{x}} \right] \cr & f'\left( x \right) = 2 - \frac{{18}}{{{x^2}}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & 2 - \frac{{18}}{{{x^2}}} = 0 \cr & {x^2} = 9 \cr & x = - 3,{\text{ }}x = - 3 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {2 - \frac{{18}}{{{x^2}}}} \right] \cr & f'\left( x \right) = \frac{{36}}{{{x^3}}} \cr & \cr & {\text{Evaluate the second derivative at }}x = - 3{\text{ and }}x = 3 \cr & *f''\left( { - 3} \right) = \frac{{36}}{{{{\left( { - 3} \right)}^3}}} < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative maximum at }}\left( { - 3,f\left( { - 3} \right)} \right) \cr & f\left( { - 3} \right) = 2\left( { - 3} \right) + \frac{{18}}{{ - 3}} = - 12 \cr & {\text{Relative maximum at }}\left( { - 3, - 12} \right) \cr & *f''\left( 3 \right) = \frac{{36}}{{{{\left( 3 \right)}^3}}} > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( {3,f\left( 3 \right)} \right) \cr & f\left( 3 \right) = 2\left( 3 \right) + \frac{{18}}{3} = 12 \cr & {\text{Relative minimum at }}\left( {3,12} \right) \cr} $$