Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 45

$$\eqalign{ & \left( {\text{a}} \right)x = \frac{{\sqrt 2 }}{2} \cr & \left( {\text{b}} \right){\text{absolute maximum of }}1 + \pi {\text{ at }}x = - 1 \cr & {\text{absolute minimum of 1 at }}x = 1 \cr & \left( {\text{c}} \right){\text{ graph}} \cr} $$

$$\eqalign{ & f\left( x \right) = {x^2} + {\cos ^{ - 1}}x{\text{ on }}\left[ { - 1,1} \right] \cr & \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} + {{\cos }^{ - 1}}x} \right] \cr & f'\left( x \right) = 2x - \frac{1}{{\sqrt {1 - {x^2}} }} \cr & \cr & \left( a \right){\text{ Set the derivative equals to 0}} \cr & f'\left( x \right) = 0 \cr & 2x - \frac{1}{{\sqrt {1 - {x^2}} }} = 0 \cr & 2x = \frac{1}{{\sqrt {1 - {x^2}} }} \cr & {\text{Square both sides}} \cr & 4{x^2} = \frac{1}{{1 - {x^2}}} \cr & 4{x^2} - 4{x^4} = 1 \cr & 4{x^4} - 4{x^2} + 1 = 0 \cr & {\text{Factoring}} \cr & {\left( {2{x^2} - 1} \right)^2} = 0 \cr & 2{x^2} - 1 = 0 \cr & \underbrace {x = - \frac{{\sqrt 2 }}{2}}_{{\text{Extraneous solution}}}{\text{ and }}x = \frac{{\sqrt 2 }}{2} \cr & {\text{And the derivative is not defined at }}x = \pm 1,{\text{ but they are}} \cr & {\text{the endpoints}}{\text{, then the critical point is:}} \cr & x = \frac{{\sqrt 2 }}{2} \cr & \cr & \left( {\text{b}} \right){\text{Evaluate }}f\left( x \right){\text{ at the critical points and the endpoints }}\left[ { - 1,1} \right] \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^2} + {\cos ^{ - 1}}\left( { - 1} \right) = 1 + \pi \cr & f\left( {\frac{{\sqrt 2 }}{2}} \right) = {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} + {\cos ^{ - 1}}\left( {\frac{{\sqrt 2 }}{2}} \right) = \frac{1}{2} + \frac{\pi }{4} \cr & f\left( 1 \right) = {\left( 1 \right)^2} + {\cos ^{ - 1}}\left( 1 \right) = 1 \cr & {\text{*The largest of these function values is }}f\left( { - 1} \right) = 1 + \pi ,{\text{ then}} \cr & {\text{the absolute maximum of }}f\left( x \right){\text{ on }}\left[ { - 1,1} \right]{\text{ is }}1 + \pi \cr & {\text{*The smallest of these function values is }}f\left( 1 \right) = 1,{\text{ then}} \cr & {\text{the absolute minimum of }}f\left( x \right){\text{ on }}\left[ { - 1,1} \right]{\text{ 1}} \cr & {\text{*From the graph we can see that the critical point }}x = \frac{{\sqrt 2 }}{2}{\text{ }} \cr & {\text{corresponds to neither a local maximum nor a local minimum}}{\text{.}} \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$