Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 287: 38

$t = 0$

$$\eqalign{ & q\left( t \right) = \frac{{{t^2} + 9}}{{{t^2} - 9}} \cr & {\text{The domain of the function is}} \cr & D = \left\{ {\left. {t \in R} \right|t \ne \pm 3} \right\} \cr & {\text{Differentiate with respect to }}t \cr & q\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{{t^2} + 9}}{{{t^2} - 9}}} \right] \cr & {\text{Using the quotient rule for derivatives}} \cr & q'\left( t \right) = \frac{{\left( {{t^2} - 9} \right)\frac{d}{{dt}}\left[ {{t^2} + 9} \right] - \left( {{t^2} + 9} \right)\frac{d}{{dt}}\left[ {{t^2} - 9} \right]}}{{{{\left( {{t^2} - 9} \right)}^2}}} \cr & {\text{Computing derivatives and simplify}} \cr & q'\left( t \right) = \frac{{\left( {{t^2} - 9} \right)\left( {2t} \right) - \left( {{t^2} + 9} \right)\left( {2t} \right)}}{{{{\left( {{t^2} - 9} \right)}^2}}} \cr & q'\left( t \right) = \frac{{2{t^3} - 18t - 2{t^3} - 18t}}{{{{\left( {{t^2} - 9} \right)}^2}}} \cr & q'\left( t \right) = \frac{{ - 36t}}{{{{\left( {{t^2} - 9} \right)}^2}}} \cr & {\text{The derivative does not exist when the denominator is 0}} \cr & {\left( {{t^2} - 9} \right)^2} = 0 \to t = \pm 3 \cr & t = \pm 3{\text{ are not in the domain of the given function}}{\text{, hence}} \cr & {\text{they are not critical points}}{\text{.}} \cr & {\text{Find the values of }}t{\text{ where }}q'\left( t \right) = 0 \cr & \frac{{ - 36t}}{{{{\left( {{t^2} - 9} \right)}^2}}} = 0 \cr & - 36t = 0 \cr & t = 0 \cr & \cr & {\text{Applying the definition of critical numbers:}} \cr & {\text{A critical number of a function is a number }}c{\text{ in the domain }} \cr & {\text{off such that either }}f'\left( c \right) = 0{\text{ or }}f'\left( c \right){\text{ does not exist}}{\text{, so the}} \cr & {\text{critical number is:}} \cr & t = 0 \cr} $$