Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 29

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

$$\eqalign{ & f\left( x \right) = \frac{x}{{{x^2} + 1}} \cr & {\text{The domain of the funcion is all real numbers}}{\text{, so its derivative exists }} \cr & {\text{everywhere}}{\text{.}} \cr & f'\left( x \right) = \frac{{\left( {{x^2} + 1} \right)\left( 1 \right) - x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{{x^2} + 1 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & \frac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0 \cr & 1 - {x^2} = 0 \cr & \left( {1 + x} \right)\left( {1 - x} \right) = 0 \cr & {\text{Solving this equation gives the critical points}} \cr & x = - 1\,{\text{ and }}x = 1 \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 1} \right) = \frac{{ - 1}}{{{{\left( { - 1} \right)}^2} + 1}} = - \frac{1}{2} \cr & f\left( 1 \right) = \frac{1}{{{{\left( 1 \right)}^2} + 1}} = \frac{1}{2} \cr & \cr & {\text{The largest of these function values is }}f\left( 1 \right) = \frac{1}{2},{\text{ which is the}} \cr & {\text{local maximum for the domain of the function}}{\text{.}} \cr & {\text{The largest of these function values is }}f\left( { - 1} \right) = - \frac{1}{2},{\text{ which is the}} \cr & {\text{local minimum for the domain of the function}}{\text{.}} \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$