Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 74

$$\eqalign{ & {\text{Relative maximum at: }}\left( {3,\frac{1}{{\sqrt {2\pi } }}} \right) \cr & {\text{Inflection points at: }}\left( {2,\frac{1}{{\sqrt {2\pi } }}{e^{ - 1/2}}} \right),{\text{ }}\left( {4,\frac{1}{{\sqrt {2\pi } }}{e^{ - 1/2}}} \right) \cr} $$

$$\eqalign{ & g\left( x \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} \cr & {\text{Calculate the first derivative}} \cr & g'\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\frac{d}{{dx}}\left[ {{e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}}} \right] \cr & g'\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\left[ { - \frac{{2\left( {x - 3} \right)}}{2}{e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}}} \right] \cr & g'\left( x \right) = - \frac{1}{{\sqrt {2\pi } }}\left( {x - 3} \right){e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} \cr & {\text{Set }}g'\left( x \right) = 0 \cr & - \frac{1}{{\sqrt {2\pi } }}\left( {x - 3} \right){e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} = 0 \cr & x - 3 = 0 \cr & x = 3 \cr & \cr & *{\text{Calculate the second derivative}} \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{1}{{\sqrt {2\pi } }}\left( {x - 3} \right){e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}}} \right] \cr & g''\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\left( {x - 3} \right)\left( {x - 3} \right){e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} - \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} \cr & \cr & {\text{Evaluate the second derivative at }}x = 3 \cr & g''\left( 0 \right) = \frac{1}{{\sqrt {2\pi } }}\left( {3 - 3} \right)\left( {3 - 3} \right){e^{ - \frac{{{{\left( {3 - 3} \right)}^2}}}{2}}} - \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {3 - 3} \right)}^2}}}{2}}} \cr & g''\left( 0 \right) = - \frac{1}{{\sqrt {2\pi } }} < 0,{\text{ Then by the second derivative test }} \cr & g\left( x \right){\text{ has a relative maximum at }}\left( {3,f\left( 3 \right)} \right) \cr & g\left( 3 \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {3 - 3} \right)}^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }} \cr & {\text{Relative maximum at }}\left( {3,\frac{1}{{\sqrt {2\pi } }}} \right) \cr & {\text{Set }}g''\left( x \right) = 0 \cr & \frac{1}{{\sqrt {2\pi } }}\left( {x - 3} \right)\left( {x - 3} \right){e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} - \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}} = 0 \cr & \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - 3} \right)}^2}}}{2}}}\left[ {\left( {x - 3} \right)\left( {x - 3} \right) - 1} \right] = 0 \cr & {x^2} - 6x + 9 - 1 = 0 \cr & {x^2} - 6x + 8 = 0 \cr & \left( {x - 4} \right)\left( {x - 2} \right) = 0 \cr & {x_1} = 2,{\text{ }}{x_2} = 4 \cr & {\text{Inflection points at }}\left( {2,g\left( 2 \right)} \right),\left( {4,g\left( 4 \right)} \right) \cr & g\left( 2 \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {2 - 3} \right)}^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}{e^{ - 1/2}} \to \left( {2,\frac{1}{{\sqrt {2\pi } }}{e^{ - 1/2}}} \right) \cr & g\left( 4 \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {4 - 3} \right)}^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}{e^{ - 1/2}} \to \left( {4,\frac{1}{{\sqrt {2\pi } }}{e^{ - 1/2}}} \right) \cr & {\text{Graph}} \cr} $$