Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 38

$${\text{Relative minimum at }}\left( {0, - \frac{{{e^{ - 1}} + e}}{2}} \right)$$

$$\eqalign{ & f\left( x \right) = x\sinh \left( {x - 1} \right) - \cosh \left( {x - 1} \right) \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x\sinh \left( {x - 1} \right) - \cosh \left( {x - 1} \right)} \right] \cr & f'\left( x \right) = x\cosh \left( {x - 1} \right) + \sinh \left( {x - 1} \right) - \sinh \left( {x - 1} \right) \cr & f'\left( x \right) = x\cosh \left( {x - 1} \right) \cr & {\text{Set }}f'\left( x \right) = 0 \cr & x\cosh \left( {x - 1} \right) = 0 \cr & \cosh \left( {x - 1} \right){\text{ is always positive, then}} \cr & x = 0 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {x\cosh \left( {x - 1} \right)} \right] \cr & f''\left( x \right) = x\sinh \left( {x - 1} \right) + \cosh \left( {x - 1} \right) \cr & \cr & {\text{Evaluate the second derivative at }}x = 0 \cr & *f''\left( 0 \right) = \left( 0 \right)\sinh \left( {0 - 1} \right) + \cosh \left( {0 - 1} \right) \cr & f''\left( 0 \right) = \cosh \left( { - 1} \right) > 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( {0,f\left( 0 \right)} \right) \cr & f\left( 0 \right) = 0\sinh \left( {0 - 1} \right) - \cosh \left( {0 - 1} \right) \cr & f\left( 0 \right) = - \cosh \left( { - 1} \right) \cr & f\left( 0 \right) = - \frac{{{e^{ - 1}} + {e^{ - \left( { - 1} \right)}}}}{2} \cr & f\left( 0 \right) = - \frac{{{e^{ - 1}} + e}}{2} \cr & {\text{Relative minimum at }}\left( {0, - \frac{{{e^{ - 1}} + e}}{2}} \right) \cr & \cr & {\text{Graph}} \cr} $$