Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 74

$$\eqalign{ & {\text{Domain: }}\left[ {0,2} \right] \cr & {\text{Range: }}\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \cr & {\text{Maximum }}\left( {2,\frac{\pi }{2}} \right){\text{ and minimum }}\left( {0, - \frac{\pi }{2}} \right) \cr & {\text{Inflection point at }}\left( {1,0} \right) \cr & {\text{Horizontal asymptotes: }}y = \pm \pi \cr} $$

$$\eqalign{ & f\left( x \right) = \arctan x + \frac{\pi }{2} \cr & {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr & \cr & {\text{*Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\arctan x + \frac{\pi }{2}} \right] \cr & f'\left( x \right) = \frac{1}{{1 + {x^2}}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \frac{1}{{1 + {x^2}}} = 0 \cr & {\text{No solution}} \cr & \cr & {\text{*Find }}f''\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{1 + {x^2}}}} \right] \cr & f''\left( x \right) = \frac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{Let }}f''\left( x \right) = 0 \cr & \frac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = 0 \cr & x = 0 \cr & {\text{Evaluate }}f''\left( 0 \right) \cr & f\left( 0 \right) = \arctan \left( 0 \right) + \frac{\pi }{2} \cr & f\left( 0 \right) = \frac{\pi }{2} \cr & {\text{Inflection point at }}\left( {0,\frac{\pi }{2}} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes: the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\arctan x + \frac{\pi }{2}} \right) = \pi \cr & \mathop {\lim }\limits_{x \to - \infty } {\left( {x - 1} \right)^5} = - \pi \cr & {\text{Horizontal asymptotes: }}y = \pm \pi \cr & \cr & {\text{Graph}} \cr} $$