Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 46

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - 1,\,\,0,\,\,1 \cr & \left( b \right){\text{Local maximum at }}x = 0 \cr & \left( c \right){\text{No absolute maximum or minimum}}{\text{.}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{{x^2} - 1}}{\text{ on the interval }}\left[ { - 4,4} \right] \cr & {\text{Derivative}} \cr & f'\left( x \right) = \frac{{\left( {{x^2} - 1} \right)\left( {2x} \right) - {x^2}\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^3} - 2x - 2{x^3}}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - 2x}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & {\text{Set the derivative to 0}} \cr & - 2x = 0 \cr & x = 0 \cr & {\text{And the derivative is not defined for }}x = \pm 1.{\text{ Then}} \cr & {\text{The critical points are:}} \cr & x = 0,\,\,x = - 1,\,\,x = 1 \cr & \cr & {\text{Evaluating the critical points and the endpoints we obtain}} \cr & f\left( { - 4} \right) = \frac{{{{\left( { - 4} \right)}^2}}}{{{{\left( { - 4} \right)}^2} - 1}} = \frac{{16}}{{15}} \cr & f\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^2}}}{{{{\left( { - 1} \right)}^2} - 1}}{\text{ Undefined}} \cr & f\left( 0 \right) = \frac{{{{\left( 0 \right)}^2}}}{{{{\left( 0 \right)}^2} - 1}} = 0 \cr & f\left( 1 \right) = \frac{{{{\left( 1 \right)}^2}}}{{{{\left( 1 \right)}^2} - 1}}{\text{ Undefined}} \cr & \cr & \mathop {\lim }\limits_{x \to {0^ - }} f'\left( x \right) = + {\text{ and }}\mathop {\lim }\limits_{x \to {0^ - }} f'\left( x \right) = - .{\text{ There is a Local maximum}}{\text{.}} \cr & {\text{at }}x = 0 \cr & \cr & {\text{Then for }}\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right){\text{ we obtain}} \cr & \mathop {\lim }\limits_{x \to - {1^ - }} = \frac{{{{\left( { - 1.01} \right)}^2}}}{{{{\left( { - 1.01} \right)}^2} - 1}} = + \to + \infty \cr & \mathop {\lim }\limits_{x \to - {1^ + }} = \frac{{{{\left( { - 0.99} \right)}^2}}}{{{{\left( { - 0.99} \right)}^2} - 1}} = + \to - \infty \cr & {\text{We can conlude that there are no exist }} \cr & {\text{abolute maximun or minimum values}}{\text{.}} \cr & \cr & {\text{Then}}{\text{, }} \cr & {\text{The largest result is 68 at }}x = - 2\,\,\,\left( {{\text{Absolute maximum}}} \right) \cr & {\text{The smallest result is }} - 3964{\text{ at }}x = 4\,\,\left( {{\text{Absolute minimum}}} \right) \cr & {\text{There are no Local min or max}} \cr & \cr & {\text{We can conclude that}} \cr & \left( a \right){\text{Critical points }}x = - 1,\,\,0,\,\,1 \cr & \left( b \right){\text{Local maximum at }}x = 0 \cr & \left( c \right){\text{No absolute maximum or minimum}}{\text{.}} \cr} $$