Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 34

$$\eqalign{ & \left( a \right){\text{Critical points }}x = 0,\,\,{\text{and}}\,\,x = \pm 1 \cr & \left( b \right)f\left( 0 \right) = 0{\text{ Is the local maximum}} \cr & \,\,\,\,\,\,f\left( 1 \right),\,\,f\left( { - 1} \right) = 1 - 2\ln 2{\text{ Is the local }}\left( {{\text{and absolute}}} \right){\text{minimum}} \cr} $$

$$\eqalign{ & f\left( x \right) = {x^2} - 2\ln \left( {{x^2} + 1} \right) \cr & {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr & {\text{Differentiate}} \cr & f'\left( x \right) = 2x - 2\left( {\frac{{2x}}{{{x^2} + 1}}} \right) \cr & f'\left( x \right) = 2x - \frac{{4x}}{{{x^2} + 1}} \cr & f'\left( x \right) = \frac{{2{x^3} + 2x - 4x}}{{{x^2} + 1}} \cr & f'\left( x \right) = \frac{{2{x^3} - 2x}}{{{x^2} + 1}} \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & 2{x^3} - 2x = 0 \cr & 2x\left( {{x^2} - 1} \right) \cr & {\text{Solving this equation gives the critical points}} \cr & x = 0,\,\,\,\,x = - 1\,{\text{ and }}x = 1 \cr & \cr & {\text{Evaluating these points}} \cr & f\left( 0 \right) = {\left( 0 \right)^2} - 2\ln \left( {{{\left( 0 \right)}^2} + 1} \right) = 0 \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^2} - 2\ln \left( {{{\left( { - 1} \right)}^2} + 1} \right) = 1 - 2\ln 2 \cr & f\left( 1 \right) = {\left( 1 \right)^2} - 2\ln \left( {{{\left( 1 \right)}^2} + 1} \right) = 1 - 2\ln 2 \cr & \cr & {\text{The largest result evaluating the critical points is }}f\left( 0 \right) = 0,{\text{ which is the}} \cr & {\text{local maximum}} \cr & \cr & {\text{We obtain the smallest results evaluating the critical points }}x = \pm 1,{\text{ }} \cr & {\text{which is the local }}\left( {{\text{and absolute}}} \right){\text{minimum}} \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$