Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 71

$${\text{Relative minimum at }}\left( {0,\frac{1}{2}} \right),{\text{ No inflection points}}{\text{.}}$$

$$\eqalign{ & f\left( x \right) = \frac{{{e^x} + {e^{ - x}}}}{2} \cr & {\text{Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{2}} \right] \cr & f'\left( x \right) = \frac{{{e^x} - {e^{ - x}}}}{2} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & \frac{{{e^x} - {e^{ - x}}}}{2} = 0 \cr & {e^x} - {e^{ - x}} = 0 \cr & x = 0 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{e^x} - {e^{ - x}}}}{2}} \right] \cr & f''\left( x \right) = \frac{{{e^x} + {e^{ - x}}}}{2} \cr & \cr & {\text{Evaluate the second derivative at }}x = 0 \cr & f''\left( 0 \right) = \frac{{{e^0} + {e^{ - 0}}}}{2} \cr & f''\left( 0 \right) = \frac{1}{2} > 0,{\text{ Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( {0,f\left( 0 \right)} \right) \cr & f\left( 0 \right) = \frac{{{e^0} + {e^{ - 0}}}}{2} = \frac{1}{2} \cr & {\text{Relative minimum at }}\left( {0,\frac{1}{2}} \right) \cr & {\text{Set }}f''\left( x \right) = 0 \cr & \frac{{{e^x} + {e^{ - x}}}}{2} = 0 \cr & {e^x} + {e^{ - x}} = 0 \cr & {\text{No real solutions, then there are no inflection points}} \cr & \cr & {\text{Graph}} \cr} $$