Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 35

$$\eqalign{ & {\text{Domain }}\left( { - \infty - \infty } \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{local minimum at }}\left( { - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt {2e} }}} \right) \cr & {\text{local maximum at }}\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt {2e} }}} \right) \cr & {\text{Inflection points: }}\left( { - \frac{{\sqrt 3 }}{2}, - \frac{{\sqrt 3 }}{2}{e^{ - \frac{3}{4}}}} \right){\text{, }}\left( {0,0} \right){\text{ and }}\left( {\frac{{\sqrt 3 }}{2},\frac{{\sqrt 3 }}{2}{e^{ - \frac{3}{4}}}} \right) \cr & {\text{No vertical asymptote}} \cr & {\text{Horizontal asymptote }}y = 0 \cr} $$

$$\eqalign{ & p\left( x \right) = x{e^{ - {x^2}}} \cr & \cr & {\text{Domain }}\left( { - \infty ,\infty } \right) \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & p\left( 0 \right) = \left( 0 \right){e^{ - {{\left( 0 \right)}^2}}} \cr & p\left( 0 \right) = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}p\left( x \right) = 0 \cr & x{e^{ - {x^2}}} = 0 \cr & {e^{ - {x^2}}} > 0{\text{ for all real number }}x,{\text{ then}} \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & p'\left( x \right) = \frac{d}{{dx}}\left[ {x{e^{ - {x^2}}}} \right] \cr & p'\left( x \right) = {e^{ - {x^2}}} + x\left( { - 2x{e^{ - {x^2}}}} \right) \cr & p'\left( x \right) = {e^{ - {x^2}}} - 2{x^2}{e^{ - {x^2}}} \cr & p'\left( x \right) = {e^{ - {x^2}}}\left( {1 - 2{x^2}} \right) \cr & \cr & {\text{Let }}p'\left( x \right) = 0{\text{ to find critical points}} \cr & {e^{ - {x^2}}}\left( {1 - 2{x^2}} \right) = 0 \cr & {e^{ - {x^2}}} > 0{\text{ for all real number }}x,{\text{ then}} \cr & 1 - 2{x^2} = 0 \cr & 2{x^2} = 1 \cr & x = \pm \frac{1}{{\sqrt 2 }} \cr & \cr & *{\text{Find the second derivative}} \cr & p''\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - {x^2}}}\left( {1 - 2{x^2}} \right)} \right] \cr & p''\left( x \right) = {e^{ - {x^2}}}\left( { - 4x} \right) + \left( {1 - 2{x^2}} \right)\left( { - 2x{e^{ - {x^2}}}} \right) \cr & p''\left( x \right) = - 2x{e^{ - {x^2}}}\left[ {2 + \left( {1 - 2{x^2}} \right)} \right] \cr & p''\left( x \right) = - 2x{e^{ - {x^2}}}\left( {3 - 2{x^2}} \right) \cr & \cr & {\text{Evaluate }}p''\left( x \right){\text{ at the critical points }}x = \pm \frac{1}{{\sqrt 2 }} \cr & p''\left( { - \frac{1}{{\sqrt 2 }}} \right) = - 2\left( { - \frac{1}{{\sqrt 2 }}} \right){e^{ - {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2}}}\left( {3 - 2{{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2}} \right) > 0 \cr & {\text{There is a local minimum at }}\left( { - \frac{1}{{\sqrt 2 }},p\left( { - \frac{1}{{\sqrt 2 }}} \right)} \right) \cr & p\left( { - \frac{1}{{\sqrt 2 }}} \right) = - \frac{1}{{\sqrt 2 }}{e^{ - {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2}}} = - \frac{1}{{\sqrt 2 }}{e^{ - \frac{1}{2}}} = - \frac{1}{{\sqrt {2e} }} \cr & \to {\text{local minimum at }}\left( { - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt {2e} }}} \right) \cr & and \cr & p''\left( {\frac{1}{{\sqrt 2 }}} \right) = - 2\left( {\frac{1}{{\sqrt 2 }}} \right){e^{ - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}\left( {3 - 2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right) < 0 \cr & {\text{There is a local maximum at }}\left( {\frac{1}{{\sqrt 2 }},p\left( {\frac{1}{{\sqrt 2 }}} \right)} \right) \cr & p\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{1}{{\sqrt 2 }}{e^{ - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} = \frac{1}{{\sqrt 2 }}{e^{ - \frac{1}{2}}} = \frac{1}{{\sqrt {2e} }} \cr & \to {\text{local maximum at }}\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt {2e} }}} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}p''\left( x \right) = 0 \cr & - 2x{e^{ - {x^2}}}\left( {3 - 2{x^2}} \right) = 0 \cr & - 2x = 0 \to x = 0 \cr & 3 - 2{x^2} = 0 \to x = \pm \sqrt {\frac{3}{2}} ,{\text{ then}} \cr & p\left( { - \frac{{\sqrt 3 }}{2}} \right) = \left( { - \frac{{\sqrt 3 }}{2}} \right){e^{ - {{\left( { - \frac{{\sqrt 3 }}{2}} \right)}^2}}} = - \frac{{\sqrt 3 }}{2}{e^{ - \frac{3}{4}}} \cr & p\left( 0 \right) = \left( 0 \right){e^{ - {{\left( 0 \right)}^2}}} = 0 \cr & p\left( {\frac{{\sqrt 3 }}{2}} \right) = \left( {\frac{{\sqrt 3 }}{2}} \right){e^{ - {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{{\sqrt 3 }}{2}{e^{ - \frac{3}{4}}} \cr & {\text{The inflection points are: }} \cr & \left( { - \frac{{\sqrt 3 }}{2}, - \frac{{\sqrt 3 }}{2}{e^{ - \frac{3}{4}}}} \right){\text{, }}\left( {0,0} \right){\text{ and }}\left( {\frac{{\sqrt 3 }}{2},\frac{{\sqrt 3 }}{2}{e^{ - \frac{3}{4}}}} \right) \cr & \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {x{e^{ - x}}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {x{e^{ - x}}} \right) = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr & {\text{Graph}} \cr} $$