Chapter 3 - Applications of Differentiation - Review Exercises - Page 238: 30

$$\eqalign{ & \left( {\text{a}} \right)x = \pm \sqrt {\frac{8}{3}} \cr & \left( {\text{b}} \right){\text{ Increasing on: }}\left( { - \infty , - \sqrt {\frac{8}{3}} } \right),{\text{ }}\left( {\sqrt {\frac{8}{3}} ,\infty } \right) \cr & {\text{Decreasing on: }}\left( { - \sqrt {\frac{8}{3}} ,\sqrt {\frac{8}{3}} } \right) \cr & \left( {\text{c}} \right){\text{relative maximum at }}\left( { - \sqrt {\frac{8}{3}} ,\frac{{8\sqrt 6 }}{9}} \right) \cr & {\text{relative minimum at }}\left( {\sqrt {\frac{8}{3}} , - \frac{{8\sqrt 6 }}{9}} \right) \cr} $$

$$\eqalign{ & g\left( x \right) = \frac{{{x^3} - 8x}}{4} \cr & g\left( x \right) = \frac{{{x^3}}}{4} - 2x \cr & \cr & \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^3}}}{4} - 2x} \right] \cr & g'\left( x \right) = \frac{{3{x^2}}}{4} - 2 \cr & {\text{Calculating the critical points, set }}g'\left( x \right) = 0 \cr & g'\left( x \right) = 0 \cr & \frac{{3{x^2}}}{4} - 2 = 0 \cr & 3{x^2} = 8 \cr & x = \pm \sqrt {\frac{8}{3}} \cr & \cr & \left( {\text{b}} \right){\text{Set the intervals }}\left( { - \infty , - \sqrt {\frac{8}{3}} } \right),\left( { - \sqrt {\frac{8}{3}} ,\sqrt {\frac{8}{3}} } \right),\left( {\sqrt {\frac{8}{3}} ,\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of g}}'\left( x \right)}&{{\text{Conclusion}}} \\ {\left( { - \infty , - \sqrt {\frac{8}{3}} } \right)}&{x = - 2}&{1 > 0}&{{\text{Increasing}}} \\ {\left( { - \sqrt {\frac{8}{3}} ,\sqrt {\frac{8}{3}} } \right)}&{x = 0}&{ - 2 < 0}&{{\text{Decreasing}}} \\ {\left( {\sqrt {\frac{8}{3}} ,\infty } \right)}&{x = 2}&{1 > 0}&{{\text{Increasing}}} \end{array}}\] $$\eqalign{ & \left( {\text{c}} \right){\text{ By Theorem 3}}{\text{.6}} \cr & *g'\left( x \right){\text{changes from positive to negative at }}x = - \sqrt {\frac{8}{3}} ,{\text{ so }}g\left( x \right) \cr & {\text{has a relative maximum at }}\left( { - \sqrt {\frac{8}{3}} ,f\left( { - \sqrt {\frac{8}{3}} } \right)} \right) \cr & g\left( { - \sqrt {\frac{8}{3}} } \right) = \frac{1}{4}{\left( { - \sqrt {\frac{8}{3}} } \right)^3} - 2\left( { - \sqrt {\frac{8}{3}} } \right) = \frac{{8\sqrt 6 }}{9} \cr & *g'\left( x \right){\text{changes from negative to positive at }}x = \sqrt {\frac{8}{3}} ,{\text{ so }}g\left( x \right) \cr & {\text{has a relative minimum at }}\left( {\sqrt {\frac{8}{3}} ,g\left( {\sqrt {\frac{8}{3}} } \right)} \right) \cr & g\left( {\sqrt {\frac{8}{3}} } \right) = \frac{1}{4}{\left( {\sqrt {\frac{8}{3}} } \right)^3} - 2\left( {\sqrt {\frac{8}{3}} } \right) = - \frac{{8\sqrt 6 }}{9} \cr & \cr & \left( {\text{d}} \right){\text{Graph}} \cr} $$