Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 69

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr }$$

$$\eqalign{ & f\left( x \right) = x\sqrt {16 - {x^2}} \cr & {\text{Domain:}}16 - {x^2} \geqslant 0 \cr & - 4 \leqslant x \leqslant 4 \cr & {\text{Domain: }}\left[ { - 4,4} \right] \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = 0\sqrt {16 - {0^2}} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = x\sqrt {16 - {x^2}} \cr & x = - 4,{\text{ }}x = 0,{\text{ }}x = 4 \cr & x{\text{ - intercept }}\left( { - 4,0} \right),\left( {0,0} \right),\left( {4,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {x\sqrt {16 - {x^2}} } \right] \cr & y' = x\frac{d}{{dx}}\left[ {\sqrt {16 - {x^2}} } \right] + \sqrt {16 - {x^2}} \frac{d}{{dx}}\left[ x \right] \cr & y' = x\left( {\frac{{ - 2x}}{{2\sqrt {16 - {x^2}} }}} \right) + \sqrt {16 - {x^2}} \left( 1 \right) \cr & y' = - \frac{{{x^2}}}{{\sqrt {16 - {x^2}} }} + \sqrt {16 - {x^2}} \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & - \frac{{{x^2}}}{{\sqrt {16 - {x^2}} }} + \sqrt {16 - {x^2}} = 0 \cr & \sqrt {16 - {x^2}} = \frac{{{x^2}}}{{\sqrt {16 - {x^2}} }} \cr & 16 - {x^2} = {x^2} \cr & 2{x^2} = 16 \cr & {x^2} = 8 \cr & x = \pm 2\sqrt 2 \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ { - \frac{{{x^2}}}{{\sqrt {16 - {x^2}} }} + \sqrt {16 - {x^2}} } \right] \cr & {\text{Using a CAS we obtain }} \cr & y'' = \frac{{2x\left( {{x^2} - 24} \right)}}{{{{\left( {16 - {x^2}} \right)}^{3/2}}}} \cr & {\text{Evaluate }}y''\left( x \right){\text{ at }}x = \pm 2\sqrt 2 \cr & y''\left( { - 2\sqrt 2 } \right) = 4 > 0,{\text{ relative minimum at }}x = - 2\sqrt 2 \cr & f\left( {2\sqrt 2 } \right) = - 8 \to {\text{Relative minimum at }}\left( { - 2\sqrt 2 , - 8} \right) \cr & {\text{Evaluate }}y''\left( x \right){\text{ at }}x = 2\sqrt 2 \cr & y''\left( {2\sqrt 2 } \right) = - 4 < 0,{\text{ relative maximum at }}x = 2\sqrt 2 \cr & f\left( {2\sqrt 2 } \right) = 8 \to {\text{Relative maximum at }}\left( {2\sqrt 2 ,8} \right) \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & \frac{{2x\left( {{x^2} - 24} \right)}}{{{{\left( {16 - {x^2}} \right)}^{3/2}}}} = 0 \cr & {x^2} - 24 = 0,{\text{ }}x = 0 \cr & x = \pm \sqrt {24} \left( {{\text{These vales are not in the domain}}} \right),{\text{ }}x = 0 \cr & x = 0 \cr & {\text{Inflection point }}\left( {0,f\left( 0 \right)} \right) \to \left( {0,0} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes, the denominator is 1}}{\text{.}} \cr & \left( { - \infty ,{\text{ or }}\infty } \right){\text{ are not in the domain, so:}} \cr & {\text{No horizontal asymptotes}}{\text{.}} \cr & \cr & {\text{Graph}} \cr} $$