Chapter 3 - Applications of Differentiation - 3.3 Exercises - Page 183: 36

$$\eqalign{ & \left( {\text{a}} \right)x = - 3,{\text{ }}x = - 1,{\text{ }}x = 1 \cr & \left( {\text{b}} \right){\text{Decreasing on: }}\left( { - 3, - 1} \right),\left( { - 1,1} \right) \cr & {\text{Increasing on: }}\left( { - \infty , - 3} \right){\text{,}}\left( {1,\infty } \right) \cr & \left( {\text{c}} \right){\text{relative minimum}}\left( {1,0} \right),{\text{ relative maximum}}\left( { - 3, - 8} \right) \cr} $$

$$\eqalign{ & {\text{We have }}f\left( x \right) = \frac{{{x^2} - 2x + 1}}{{x + 1}} \cr & \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2} - 2x + 1}}{{x + 1}}} \right] \cr & {\text{By the quotient rule}} \cr & f'\left( x \right) = \frac{{\left( {x + 1} \right)\left( {2x - 2} \right) - \left( {{x^2} - 2x + 1} \right)\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^2} - 2x + 2x - 2 - {x^2} + 2x - 1}}{{{{\left( {x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{{x^2} + 2x - 3}}{{{{\left( {x + 1} \right)}^2}}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & \frac{{{x^2} + 2x - 3}}{{{{\left( {x + 1} \right)}^2}}} = 0 \cr & {x^2} + 2x - 3 = 0 \cr & \left( {x + 3} \right)\left( {x - 1} \right) = 0 \cr & x = - 3,{\text{ }}x = 1 \cr & {\text{The derivative is not defined at }}x = - 1,{\text{ so we obtain the critical }} \cr & {\text{points, }}x = - 3,{\text{ }}x = - 1,{\text{ }}x = 1 \cr & \cr & \left( {\text{b}} \right) \cr & {\text{Set the intervals }}\left( { - \infty , - 3} \right),\left( { - 3, - 1} \right),\left( { - 1,1} \right),\left( {1,\infty } \right) \cr} $$ \[\begin{gathered} {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \hfill \\ \begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty , - 3} \right)}&{\left( { - 3, - 1} \right)}&{\left( { - 1,1} \right)}&{\left( {1,\infty } \right)} \\ {{\text{Test Value}}}&{x = - 4}&{x = - 2}&{x = 0}&{x = 2} \\ {{\text{Sign of }}f'\left( x \right)}&{\frac{5}{9} > 0}&{ - 3 < 0}&{ - 3 < 0}&{\frac{5}{9} > 0} \\ {{\text{Conclusion}}}&{{\text{Incr}}{\text{.}}}&{{\text{Decr}}{\text{.}}}&{{\text{Decr}}{\text{.}}}&{{\text{Incr}}{\text{.}}} \end{array} \hfill \\ \end{gathered} \] $$\eqalign{ & f'\left( x \right){\text{ changes from positive to negative at }}x = - 3,{\text{ so }}f\left( x \right) \cr & {\text{has a relative maximum at }}\left( { - 3,f\left( { - 3} \right)} \right) \cr & f\left( { - 3} \right) = \frac{{{{\left( { - 3} \right)}^2} - 2\left( { - 3} \right) + 1}}{{\left( { - 3} \right) + 1}} = - 8 \cr & f'\left( x \right){\text{ changes from negative to positive at }}x = 1,{\text{ so }}f\left( x \right) \cr & {\text{has a relative minimum at }}\left( {1,f\left( 1 \right)} \right) \cr & f\left( 1 \right) = \frac{{{{\left( 1 \right)}^2} - 2\left( 1 \right) + 1}}{{\left( 1 \right) + 1}} = 0 \cr} $$