Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 34

$$\eqalign{ & {\text{Domain }}\left( { - \infty - \infty } \right) \cr & y{\text{ - intercept }}\left( {0,1} \right) \cr & {\text{No }}x{\text{ - intercepts}} \cr & {\text{Local maximum }}\left( {0,1} \right) \cr & {\text{Inflection points: }}\left( { - 1,\frac{1}{{\sqrt e }}} \right){\text{ and }}\left( {1,\frac{1}{{\sqrt e }}} \right) \cr & {\text{No vertical asymptote}} \cr & {\text{Horizontal asymptote }}y = 0 \cr} $$

$$\eqalign{ & g\left( x \right) = {e^{ - {x^2}/2}} \cr & {\text{Domain }}\left( { - \infty ,\infty } \right) \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & g\left( 0 \right) = {e^{ - {{\left( 0 \right)}^2}/2}} \cr & g\left( 0 \right) = 1 \cr & y{\text{ - intercept }}\left( {0,1} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}g\left( x \right) = 0 \cr & {e^{ - {x^2}/2}} = 0 \cr & {e^{ - {x^2}/2}} > 0{\text{ for all real number }}x,{\text{ then}} \cr & {\text{No }}x{\text{ - intercepts}} \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - {x^2}/2}}} \right] \cr & g'\left( x \right) = {e^{ - {x^2}/2}}\frac{d}{{dx}}\left[ { - \frac{{{x^2}}}{2}} \right] \cr & g'\left( x \right) = {e^{ - {x^2}/2}}\left( { - \frac{{2x}}{2}} \right) \cr & g'\left( x \right) = - x{e^{ - {x^2}/2}} \cr & \cr & {\text{Let }}g'\left( x \right) = 0{\text{ to find critical points}} \cr & - x{e^{ - {x^2}/2}} = 0 \cr & {e^{ - {x^2}/2}} > 0{\text{ for all real number }}x,{\text{ then}} \cr & - x = 0 \cr & x = 0 \cr & \cr & *{\text{Find the second derivative}} \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ { - x{e^{ - {x^2}/2}}} \right] \cr & g''\left( x \right) = - x\left( { - x{e^{ - {x^2}/2}}} \right) - \left( 1 \right)\left( {{e^{ - {x^2}/2}}} \right) \cr & g''\left( x \right) = {x^2}{e^{ - {x^2}/2}} - {e^{ - {x^2}/2}} \cr & g''\left( x \right) = {e^{ - {x^2}/2}}\left( {{x^2} - 1} \right) \cr & {\text{Evaluate }}g''\left( x \right){\text{ at the critical point }}x = 0 \cr & g''\left( x \right) = {e^{ - {{\left( 0 \right)}^2}/2}}\left( {{{\left( 0 \right)}^2} - 1} \right) = - 1 < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}\left( {0,g\left( 0 \right)} \right) \cr & g\left( 0 \right) = {e^{ - {{\left( 0 \right)}^2}/2}} = 1 \cr & \to {\text{local maximum at }}\left( {0,1} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}g''\left( x \right) = 0 \cr & {e^{ - {x^2}/2}}\left( {{x^2} - 1} \right) = 0 \cr & {x^2} - 1 = 0 \cr & x = \pm 1 \cr & g\left( { - 1} \right) = {e^{ - {{\left( { - 1} \right)}^2}/2}} = \frac{1}{{\sqrt e }} \cr & g\left( 1 \right) = {e^{ - {{\left( 1 \right)}^2}/2}} = \frac{1}{{\sqrt e }} \cr & {\text{The inflection points are: }}\left( { - 1,\frac{1}{{\sqrt e }}} \right){\text{ and }}\left( {1,\frac{1}{{\sqrt e }}} \right) \cr & \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - {x^2}/2}}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {{e^{ - {x^2}/2}}} \right) = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr & {\text{Graph}} \cr} $$