Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 69

$$\eqalign{ & {\text{Relative minimum at }}\left( { - 1.272,3.7473} \right) \cr & {\text{Relative maximum at }}\left( {1.272, - 0.6057} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \operatorname{arcsec} x - x \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\operatorname{arcsec} x - x} \right] \cr & f'\left( x \right) = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} - 1 \cr & {\text{Set }}f'\left( x \right) = 0 \cr & \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} = 1 \cr & \left| x \right|\sqrt {{x^2} - 1} = 1 \cr & {x^2}\left( {{x^2} - 1} \right) = 1 \cr & {x^4} - {x^2} - 1 = 0 \cr & {\text{By the quadratic formula}} \cr & {x^2} = \frac{{ - \left( { - 1} \right) \pm \sqrt {\left( { - 1} \right) - 4\left( 1 \right)\left( { - 1} \right)} }}{2} \cr & {x^2} = \frac{{1 \pm \sqrt 5 }}{2} \cr & {x^2} = \frac{{1 + \sqrt 5 }}{2} \cr & x = \pm \sqrt {\frac{{1 + \sqrt 5 }}{2}} \cr & x = \pm 1.272 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} - 1} \right] \cr & f''\left( x \right) = - \frac{{x\left( {\frac{{2x}}{{2\sqrt {{x^2} - 1} }}} \right) + \sqrt {{x^2} - 1} }}{{{x^2}\left( {{x^2} - 1} \right)}} \cr & f''\left( x \right) = - \frac{{2{x^2} - {x^2} - 1}}{{2{x^2}{{\left( {{x^2} - 1} \right)}^{3/2}}}} \cr & f''\left( x \right) = - \frac{{{x^2} - 1}}{{2{x^2}{{\left( {{x^2} - 1} \right)}^{3/2}}}} \cr & f''\left( x \right) = - \frac{1}{{2{x^2}\sqrt {{x^2} - 1} }} \cr & \cr & {\text{*Evaluate the second derivative at }}x = - 1.272 \cr & f''\left( { - 1.272} \right) = - \frac{1}{{2{{\left( { - 1.272} \right)}^2}\sqrt {{{\left( { - 1.272} \right)}^2} - 1} }} = 2.84 > 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( { - 1.272,f\left( { - 1.272} \right)} \right) \cr & f\left( { - 1.272} \right) = \operatorname{arcsec} \left( { - 1.272} \right) - \left( { - 1.272} \right) \approx 3.7473 \cr & {\text{Relative minimum at }}\left( { - 1.272,3.7473} \right) \cr & *{\text{Evaluate the second derivative at }}x = 1.272 \cr & f''\left( {1.272} \right) = \frac{1}{{2{{\left( {1.272} \right)}^2}\sqrt {{{\left( {1.272} \right)}^2} - 1} }} = - 2.84 < 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( {1.272,f\left( {1.272} \right)} \right) \cr & f\left( {1.272} \right) = \operatorname{arcsec} \left( {1.272} \right) - \left( {1.272} \right) \approx - 0.6057 \cr & {\text{Relative maximum at }}\left( {1.272, - 0.6057} \right) \cr} $$