Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 22

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Local minimum at }}\left( {0,0} \right) \cr & {\text{Inflection points: }}\left( { - 1,\ln 2} \right),\left( {1,\ln 2} \right) \cr & {\text{No horizontal asymptotes}}{\text{.}} \cr & {\text{No vertical asymptotes}}{\text{.}} \cr} $$

$$\eqalign{ & f\left( x \right) = \ln \left( {{x^2} + 1} \right) \cr & {\text{ln}}\left( u \right){\text{ is defined for all }}u > 0, \cr & {\text{The domain of the function is }} \cr & {x^2} + 1 > 0 \cr & {\text{This inequation has solution for all real numbers}}{\text{, then}} \cr & {\text{The domain of }}f\left( x \right){\text{ is }}\left( { - \infty ,\infty } \right) \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \ln \left( {{0^2} + 1} \right) \cr & f\left( 0 \right) = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = \ln \left( {{x^2} + 1} \right) \cr & {e^0} = {e^{\ln \left( {{x^2} + 1} \right)}} \cr & 1 = {x^2} + 1 \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {{x^2} + 1} \right)} \right] \cr & f'\left( x \right) = \frac{{2x}}{{{x^2} + 1}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & \frac{{2x}}{{{x^2} + 1}} = 0 \cr & 2x = 0 \cr & x = 0 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x}}{{{x^2} + 1}}} \right] \cr & f''\left( x \right) = \frac{{\left( {{x^2} + 1} \right)\left( 2 \right) - 2x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{Simplifying}} \cr & f''\left( x \right) = \frac{{2{x^2} + 2 - 4{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & f''\left( x \right) = \frac{{2 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{Evaluate the second derivative at the critical point }}x = 0 \cr & f''\left( 0 \right) = \frac{{2 - 2{{\left( 0 \right)}^2}}}{{{{\left( {{0^2} + 1} \right)}^2}}} = 2 > 0 \to {\text{Local minimum at }}\left( {0,f\left( 0 \right)} \right) \cr & f\left( 0 \right) = \ln \left( {{0^2} + 1} \right) = 0 \cr & {\text{There is a local minimum at }}\left( {0,0} \right) \cr & \cr & *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{{2 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0 \cr & 2 - 2{x^2} = 0 \cr & {x^2} = 1 \cr & x = \pm 1 \cr & {\text{Inflection points at }}x = \pm 1 \cr & f\left( { - 1} \right) = \ln 2 \cr & f\left( 1 \right) = \ln 2 \cr & {\text{The inflection points are:}} \cr & \left( { - 1,\ln 2} \right),\left( {1,\ln 2} \right) \cr & \cr & {\text{The denominator is never zero}}{\text{, then there are no}} \cr & {\text{vertical asymptotes}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\ln \left( {{x^2} + 1} \right)} \right) = \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\ln \left( {{x^2} + 1} \right)} \right) = \infty \cr & {\text{No horizontal asymptotes}}{\text{.}} \cr & \cr & {\text{Graph}} \cr} $$