Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 39

$$\eqalign{ & {\text{Relative minimum at }}\left( { - 1.19967, - 0.6627} \right) \cr & {\text{Relative maximum at }}\left( {1.19967,0.6627} \right) \cr} $$

$$\eqalign{ & g\left( x \right) = x\operatorname{sech} x \cr & {\text{*Calculate the first derivative}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {x\operatorname{sech} x} \right] \cr & g'\left( x \right) = - x\operatorname{sech} x\tanh x + \operatorname{sech} x \cr & {\text{Set }}g'\left( x \right) = 0 \cr & \operatorname{sech} x\left( {1 - x\tanh x} \right) = 0 \cr & \operatorname{sech} x = 0{\text{ or }}1 - x\tanh x = 0 \cr & \operatorname{sech} x{\text{ is always positive, then}} \cr & 1 - x\tanh x = 0 \cr & x\tanh x = 1 \cr & x\left( {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right) = 1 \cr & {\text{Solving by using a scientific calculator we obtain:}} \cr & x \approx - 1.19967{\text{ and }}x \approx 1.19967 \cr & \cr & *{\text{Calculate the second derivative}} \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & g''\left( x \right) = \frac{d}{{dx}}\left[ { - x\operatorname{sech} x\tanh x + \operatorname{sech} x} \right] \cr & g''\left( x \right) = - x\left( {\operatorname{sech} x{{\operatorname{sech} }^2}x - \operatorname{sech} x{{\tanh }^2}x} \right) - \operatorname{sech} x\tanh x \cr & - \operatorname{sech} x\tanh x \cr & \cr & {\text{*Evaluate the second derivative at }}x = - 1.19967 \cr & g''\left( { - 1.19967} \right) = 0.6627 > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & g\left( x \right){\text{ has a relative minimum at }}\left( { - 1.19967,g\left( { - 1.19967} \right)} \right) \cr & g\left( { - 1.19967} \right) = - 0.6627 \cr & {\text{Relative minimum at }}\left( { - 1.19967, - 0.6627} \right) \cr & \cr & {\text{*Evaluate the second derivative at }}x = - 1.19967 \cr & g''\left( {1.19967} \right) = - 0.6627 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & g\left( x \right){\text{ has a relative maximum at }}\left( {1.19967,g\left( {1.19967} \right)} \right) \cr & g\left( {1.19967} \right) = 0.6627 \cr & {\text{Relative maximum at }}\left( {1.19967,0.6627} \right) \cr & \cr & {\text{Graph}} \cr} $$