Chapter 3 - Applications of Differentiation - 3.4 Exercises - Page 192: 39

$$\eqalign{ & {\text{Relative maximum at }}\left( { - 2, - 4} \right) \cr & {\text{Relative minimum at }}\left( {2,4} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = x + \frac{4}{x} \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x + \frac{4}{x}} \right] \cr & f'\left( x \right) = 1 - \frac{4}{{{x^2}}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & 1 - \frac{4}{{{x^2}}} = 0 \cr & {x^2} = 4 \cr & x = - 2,{\text{ }}x = 2 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {1 - \frac{4}{{{x^2}}}} \right] \cr & f''\left( x \right) = 0 - 4\left( { - 2} \right){x^{ - 3}} \cr & f''\left( x \right) = \frac{8}{{{x^3}}} \cr & \cr & {\text{Evaluate the second derivative at }}x = - 2{\text{ and }}x = 2 \cr & *f''\left( { - 2} \right) = \frac{8}{{{{\left( { - 2} \right)}^3}}} = - 1 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative maximum at }}\left( { - 2,f\left( { - 2} \right)} \right) \cr & f\left( { - 2} \right) = - 2 + \frac{4}{{ - 2}} = - 4 \cr & {\text{Relative maximum at }}\left( { - 2, - 4} \right) \cr & *f''\left( 2 \right) = \frac{8}{{{{\left( 2 \right)}^3}}} = 1 > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( {2,f\left( 2 \right)} \right) \cr & f\left( 2 \right) = 2 + \frac{4}{2} = 4 \cr & {\text{Relative minimum at }}\left( {2,4} \right) \cr} $$