Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 76

$$\eqalign{ & {\text{Domain: }}\left[ { - 4,4} \right] \cr & {\text{Range : }}\left[ {0,\pi } \right] \cr & {\text{Maximum:}}\left( { - 4,\pi } \right){\text{ and minimum:}}\left( {4,0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \arccos \frac{x}{4} \cr & {\text{The domain of the function is }}\left[ { - 4,4} \right] \cr & \cr & {\text{*Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\arccos \frac{x}{4}} \right] \cr & f'\left( x \right) = - \frac{{\left( {1/4} \right)}}{{\sqrt {1 - \left( {x/4} \right)} }} \cr & f'\left( x \right) = - \frac{1}{{2\sqrt {4 - {x^2}} }} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & - \frac{1}{{2\sqrt {4 - {x^2}} }} = 0 \cr & {\text{No solution}} \cr & \cr & {\text{*Find }}f''\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{1}{{2\sqrt {4 - {x^2}} }}} \right] \cr & f''\left( x \right) = - \frac{x}{{2{{\left( {4 - {x^2}} \right)}^{3/2}}}} \cr & {\text{Let }}f''\left( x \right) = 0 \cr & - \frac{x}{{2{{\left( {4 - {x^2}} \right)}^{3/2}}}} = 0 \cr & x = 0 \cr & {\text{Evaluate }}f''\left( 0 \right) \cr & f\left( 0 \right) = \arccos \left( 0 \right) = \frac{\pi }{2} \cr & f\left( 0 \right) = \frac{\pi }{2} \cr & {\text{Inflection point at }}\left( {0,\frac{\pi }{2}} \right) \cr & \cr & {\text{*Evaluate }}f\left( x \right){\text{ at the endpoints}} \cr & f\left( { - 4} \right) = \arccos \left( {\frac{{ - 4}}{4}} \right) = \pi \cr & f\left( 4 \right) = \arccos \left( {\frac{4}{4}} \right) = 0,{\text{ then}} \cr & {\text{Maximum:}}\left( { - 4,\pi } \right){\text{ and minimum:}}\left( {4,0} \right) \cr & \cr & {\text{With the endpoints and the domain, we conclude that}} \cr & {\text{the range is: }}\left[ {0,\pi } \right] \cr & \cr & {\text{Graph}} \cr} $$