Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 287: 37

$x = 2,{\text{ }}x = - 1$

$$\eqalign{ & p\left( x \right) = \frac{{{x^2} + 2}}{{2x - 1}} \cr & {\text{The domain of the function is}} \cr & D = \left\{ {\left. {x \in R} \right|x \ne \frac{1}{2}} \right\} \cr & {\text{Differentiate with respect to }}x \cr & p'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2} + 2}}{{2x - 1}}} \right] \cr & {\text{Using the quotient rule for derivatives}} \cr & p'\left( x \right) = \frac{{\left( {2x - 1} \right)\frac{d}{{dx}}\left[ {{x^2} + 2} \right] - \left( {{x^2} + 2} \right)\frac{d}{{dx}}\left[ {2x - 1} \right]}}{{{{\left( {2x - 1} \right)}^2}}} \cr & {\text{Computing derivatives and simplify}} \cr & p'\left( x \right) = \frac{{\left( {2x - 1} \right)\left( {2x} \right) - \left( {{x^2} + 2} \right)\left( 2 \right)}}{{{{\left( {2x - 1} \right)}^2}}} \cr & p'\left( x \right) = \frac{{4{x^2} - 2x - 2{x^2} - 4}}{{{{\left( {2x - 1} \right)}^2}}} \cr & p'\left( x \right) = \frac{{2{x^2} - 2x - 4}}{{{{\left( {2x - 1} \right)}^2}}} \cr & {\text{The derivative does not exist when the denominator is 0}} \cr & {\left( {2x - 1} \right)^2} = 0 \to x = \frac{1}{2} \cr & x = \frac{1}{2}{\text{ is not in the domain of the given function}}{\text{, hence}} \cr & {\text{it is not a critical point}}{\text{.}} \cr & {\text{Find the values of }}x{\text{ where }}p'\left( x \right) = 0 \cr & \frac{{2{x^2} - 2x - 4}}{{{{\left( {2x - 1} \right)}^2}}} = 0 \cr & 2{x^2} - 2x - 4 = 0 \cr & 2\left( {{x^2} - x - 2} \right) = 0 \cr & 2\left( {x - 2} \right)\left( {x + 1} \right) = 0 \cr & x = 2,{\text{ }}x = - 1 \cr & \cr & {\text{Applying the definition of critical numbers:}} \cr & {\text{A critical number of a function is a number }}c{\text{ in the domain }} \cr & {\text{off such that either }}f'\left( c \right) = 0{\text{ or }}f'\left( c \right){\text{ does not exist}}{\text{, so the}} \cr & {\text{critical numbers are:}} \cr & x = 2,{\text{ }}x = - 1 \cr} $$