Chapter 3 - Applications of Differentiation - 3.3 Exercises - Page 183: 33

$$\eqalign{ & \left( {\text{a}} \right)x = - \frac{1}{{\sqrt 2 }},{\text{ }}x = 0,{\text{ }}x = \frac{1}{{\sqrt 2 }} \cr & \left( {\text{b}} \right){\text{Decreasing on: }}\left( { - \frac{1}{{\sqrt 2 }},0} \right){\text{ and }}\left( {0,\frac{{\sqrt 2 }}{2}} \right),{\text{ }} \cr & {\text{Increasing on: }}\left( { - \infty , - \frac{{\sqrt 2 }}{2}} \right){\text{ and }}\left( {\frac{{\sqrt 2 }}{2},\infty } \right),{\text{ }} \cr & \left( {\text{c}} \right){\text{relative maximum}}\left( { - \frac{1}{{\sqrt 2 }}, - 2\sqrt 2 } \right) \cr & {\text{ relative minimum}}\left( {\frac{{\sqrt 2 }}{2},2\sqrt 2 } \right) \cr} $$

$$\eqalign{ & {\text{We have }}f\left( x \right) = 2x + \frac{1}{x} \cr & \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2x + \frac{1}{x}} \right] \cr & f'\left( x \right) = 2 - \frac{1}{{{x^2}}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & 2 - \frac{1}{{{x^2}}} = 0 \cr & \frac{{2{x^2} - 1}}{{{x^2}}} = 0 \cr & 2{x^2} - 1 = 0 \cr & x = \pm \frac{1}{{\sqrt 2 }} \cr & {\text{The derivative is not defined at }}x = 0,{\text{ so we obtain the critical }} \cr & {\text{points, }}x = - \frac{1}{{\sqrt 2 }},{\text{ }}x = 0,{\text{ }}x = \frac{1}{{\sqrt 2 }}. \cr & \cr & \left( {\text{b}} \right) \cr & {\text{Set the intervals }}\left( { - \infty , - \frac{1}{{\sqrt 2 }}} \right),\left( { - \frac{1}{{\sqrt 2 }},0} \right),\left( {0,\frac{1}{{\sqrt 2 }}} \right),\left( {\frac{1}{{\sqrt 2 }},\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty , - \frac{1}{{\sqrt 2 }}} \right)}&{\left( { - \frac{1}{{\sqrt 2 }},0} \right)}&{\left( {0,\frac{1}{{\sqrt 2 }}} \right)}&{\left( {\frac{1}{{\sqrt 2 }},\infty } \right)} \\ {{\text{Test Value}}}&{x = - 5}&{ - 0.1}&{0.1}&1 \\ {{\text{Sign of }}f'\left( x \right)}&{ > 0}&{ < 0}&{ < 0}&{ > 0} \\ {{\text{Conclusion}}}&{{\text{Incr}}{\text{.}}}&{{\text{Decr}}{\text{.}}}&{{\text{Decr}}{\text{.}}}&{{\text{Incr}}{\text{.}}} \end{array}\] $$\eqalign{ & f'\left( x \right){\text{ changes from positive to negative at }}x = - \frac{1}{{\sqrt 2 }},{\text{ so }}f\left( x \right) \cr & {\text{has a relative maximum at }}\left( { - \frac{1}{{\sqrt 2 }},f\left( { - \frac{1}{{\sqrt 2 }}} \right)} \right) \cr & f\left( { - \frac{1}{{\sqrt 2 }}} \right) = 2\left( { - \frac{1}{{\sqrt 2 }}} \right) - \sqrt 2 = - 2\sqrt 2 \cr & f'\left( x \right){\text{ changes from negative to positive at }}x = \frac{1}{{\sqrt 2 }},{\text{ so }}f\left( x \right) \cr & {\text{has a relative minimum at }}\left( {\frac{1}{{\sqrt 2 }},f\left( {\frac{1}{{\sqrt 2 }}} \right)} \right) \cr & f\left( {\frac{1}{{\sqrt 2 }}} \right) = 2\left( {\frac{1}{{\sqrt 2 }}} \right) + \sqrt 2 = 2\sqrt 2 \cr} $$