Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 73

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

$$\eqalign{ & f\left( x \right) = \arcsin \left( {x - 1} \right) \cr & {\text{The domain of }}\arcsin x{\text{ is }}\left[ { - 1,1} \right],\,{\text{then the domain of }} \cr & \arcsin \left( {x - 1} \right){\text{ is: }}\left[ { - 1 + 1,1 + 1} \right] \cr & {\text{Domain}}:{\text{ }}\left[ {0,2} \right] \cr & \cr & {\text{*Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\arcsin \left( {x - 1} \right)} \right] \cr & f'\left( x \right) = \frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }} = 0 \cr & {\text{No solution}} \cr & \cr & {\text{*Evaluate }}f\left( x \right){\text{ at the endpoints}} \cr & f\left( 0 \right) = \arcsin \left( { - 1} \right) = - \frac{\pi }{2} \cr & f\left( 2 \right) = \arcsin \left( {2 - 1} \right) = \frac{\pi }{2},{\text{ then}} \cr & {\text{Maximum }}\left( {2,\frac{\pi }{2}} \right){\text{ and minimum }}\left( {0, - \frac{\pi }{2}} \right) \cr & {\text{and the range of the funcion is }}\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \cr & \cr & {\text{*Find }}f''\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {{{\left[ {1 - {{\left( {x - 1} \right)}^2}} \right]}^{ - 1/2}}} \right] \cr & f''\left( x \right) = - \frac{1}{2}{\left[ {1 - {{\left( {x - 1} \right)}^2}} \right]^{ - 3/2}}\left( { - 2} \right)\left( {x - 1} \right) \cr & f''\left( x \right) = {\left[ {1 - {{\left( {x - 1} \right)}^2}} \right]^{ - 3/2}}\left( {x - 1} \right) \cr & f''\left( x \right) = \frac{{x - 1}}{{{{\left[ {1 - {{\left( {x - 1} \right)}^2}} \right]}^{3/2}}}} \cr & {\text{Let }}f''\left( x \right) = 0 \cr & \frac{{x - 1}}{{{{\left[ {1 - {{\left( {x - 1} \right)}^2}} \right]}^{3/2}}}} = 0 \cr & x = 1 \cr & {\text{Evaluate }}f''\left( 1 \right) \cr & f\left( 1 \right) = \arcsin \left( {1 - 1} \right) \cr & f\left( 1 \right) = 0 \cr & {\text{Inflection point at }}\left( {1,0} \right) \cr & \cr & {\text{Graph}} \cr} $$