Chapter 3 - Applications of Differentiation - Review Exercises - Page 238: 31

$$\eqalign{ & \left( {\text{a}} \right){\text{ }}x = - 8,{\text{ discontinuity at }}x = 0 \cr & \left( {\text{b}} \right){\text{Decreasing: }}\left( { - \infty , - 8} \right){\text{ and }}\left( {0,\infty } \right),{\text{ Increasing: }}\left( { - 8,0} \right) \cr & \left( {\text{c}} \right){\text{Relative minimum at }}\left( { - 8, - \frac{1}{{16}}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{x + 4}}{{{x^2}}} \cr & f\left( x \right) = \frac{1}{x} + \frac{4}{{{x^2}}} \cr & \cr & \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{x} + \frac{4}{{{x^2}}}} \right] \cr & f'\left( x \right) = - \frac{1}{{{x^2}}} - \frac{8}{{{x^3}}} \cr & {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr & f'\left( x \right) = 0 \cr & - \frac{1}{{{x^2}}} - \frac{8}{{{x^3}}} = 0 \cr & \frac{{x + 8}}{{{x^3}}} = 0 \cr & x = - 8 \cr & {\text{The derivative is not defined at }}x = 0,{\text{ so the critical}} \cr & {\text{points are }}x = 0{\text{ and }}x = - 8 \cr & \cr & \left( {\text{b}} \right){\text{Set the intervals }}\left( { - \infty , - 8} \right),\left( { - 8,0} \right),\left( {0,\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f'\left( x \right)}&{{\text{Conclusion}}} \\ {\left( { - \infty , - 8} \right)}&{x = - 10}&{ - \frac{1}{{500}} < 0}&{{\text{Decreasing}}} \\ {\left( { - 8,0} \right)}&{x = - 1}&{7 > 0}&{{\text{Increasing}}} \\ {\left( {0,\infty } \right)}&{x = 1}&{ - 9 < 0}&{{\text{Decreasing}}} \end{array}}\] $$\eqalign{ & \left( {\text{c}} \right){\text{ By Theorem 3}}{\text{.6}} \cr & *f'\left( x \right){\text{changes from negative to positive at }}x = - 8,{\text{ so }}f\left( x \right) \cr & {\text{has a relative minimum at }}\left( { - 8,f\left( { - 8} \right)} \right) \cr & f\left( { - 8} \right) = \frac{1}{{\left( { - 8} \right)}} + \frac{4}{{{{\left( { - 8} \right)}^2}}} = - \frac{1}{{16}} \cr & {\text{Relative minumum at }}\left( { - 8, - \frac{1}{{16}}} \right) \cr & \cr & \left( {\text{d}} \right){\text{ Graph}} \cr} $$