Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 75

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

$$\eqalign{ & f\left( x \right) = \operatorname{arcsec} 2x \cr & {\text{The domain of the function is }}\left( { - \infty , - \frac{1}{2}} \right] \cup \left[ {\frac{1}{2},\infty } \right) \cr & {\text{*Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\operatorname{arcsec} 2x} \right] \cr & f'\left( x \right) = \frac{2}{{\left| {2x} \right|\sqrt {{{\left( {2x} \right)}^2} - 1} }} \cr & f'\left( x \right) = \frac{1}{{\left| x \right|\sqrt {4{x^2} - 1} }} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \frac{1}{{\left| x \right|\sqrt {4{x^2} - 1} }} = 0 \cr & {\text{No solution}} \cr & \cr & {\text{*Evaluate }}f\left( x \right){\text{ at the endpoints}} \cr & f\left( { - \frac{1}{2}} \right) = \operatorname{arcsec} \left[ {2\left( { - \frac{1}{2}} \right)} \right] = \pi \cr & f\left( {\frac{1}{2}} \right) = \operatorname{arcsec} \left[ {2\left( {\frac{1}{2}} \right)} \right] = 0,{\text{ then}} \cr & {\text{Maximum:}}\left( { - \frac{1}{2},\pi } \right){\text{ and minimum:}}\left( {\frac{1}{2},0} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes: the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\operatorname{arcsec} 2x} \right) = \frac{\pi }{2} \cr & \mathop {\lim }\limits_{x \to - \infty } {\left( {\operatorname{arcsec} 2x} \right)^5} = \frac{\pi }{2} \cr & {\text{Horizontal asymptotes: }}y = \frac{\pi }{2} \cr & \cr & {\text{With the endpoints and the domain, we conclude that}} \cr & {\text{the range is: }}\left[ {0,\frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2},0} \right] \cr & \cr & {\text{Graph}} \cr} $$