Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 70

$$\eqalign{ & {\text{Relative minimum at }}\left( {\frac{{\sqrt 3 }}{2},\frac{\pi }{3} - \sqrt 3 } \right) \cr & {\text{Relative maximum at }}\left( { - \frac{{\sqrt 3 }}{2},\sqrt 3 - \frac{\pi }{3}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \arcsin x - 2x \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\arcsin x - 2x} \right] \cr & f'\left( x \right) = \frac{1}{{\sqrt {1 - {x^2}} }} - 2 \cr & {\text{Set }}f'\left( x \right) = 0 \cr & \frac{1}{{\sqrt {1 - {x^2}} }} = 2 \cr & \sqrt {1 - {x^2}} = \frac{1}{2} \cr & 1 - {x^2} = \frac{1}{4} \cr & {x^2} = \frac{3}{4} \cr & x = \pm \frac{{\sqrt 3 }}{2} \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\sqrt {1 - {x^2}} }} - 2} \right] \cr & f''\left( x \right) = - \frac{1}{2}{\left( {1 - {x^2}} \right)^{ - 3/2}}\left( { - 2x} \right) \cr & f''\left( x \right) = \frac{x}{{{{\left( {1 - {x^2}} \right)}^{3/2}}}} \cr & {\text{*Evaluate the second derivative at }}x = - \frac{{\sqrt 3 }}{2} \cr & f''\left( { - \frac{{\sqrt 3 }}{2}} \right) = \frac{{\left( { - \sqrt 3 /2} \right)}}{{{{\left( {1 - {{\left( { - \sqrt 3 /2} \right)}^2}} \right)}^{3/2}}}} = - 4\sqrt 3 < 0 \cr & {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative maximum at }}\left( { - \sqrt 3 /2,f\left( { - \sqrt 3 /2} \right)} \right) \cr & f\left( { - \sqrt 3 /2} \right) = \arcsin \left( { - \frac{{\sqrt 3 }}{2}} \right) - 2\left( { - \frac{{\sqrt 3 }}{2}} \right) = \sqrt 3 - \frac{\pi }{3} \cr & {\text{Relative maximum at }}\left( { - \frac{{\sqrt 3 }}{2},\sqrt 3 - \frac{\pi }{3}} \right) \cr & {\text{*Evaluate the second derivative at }}x = \frac{{\sqrt 3 }}{2} \cr & f''\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{{\left( {\sqrt 3 /2} \right)}}{{{{\left( {1 - {{\left( {\sqrt 3 /2} \right)}^2}} \right)}^{3/2}}}} = 4\sqrt 3 < 0 \cr & {\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( {\sqrt 3 /2,f\left( {\sqrt 3 /2} \right)} \right) \cr & f\left( {\sqrt 3 /2} \right) = \arcsin \left( {\frac{{\sqrt 3 }}{2}} \right) - 2\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{3} - \sqrt 3 \cr & {\text{Relative minimum at }}\left( {\frac{{\sqrt 3 }}{2},\frac{\pi }{3} - \sqrt 3 } \right) \cr} $$