Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.6 Exercises - Page 212: 15

Answer

$$\eqalign{ & {\text{Domain: }}\left( { - \infty ,4} \right] \cr & y{\text{ - intercept: }}\left( {0,0} \right) \cr & x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {4,0} \right) \cr & {\text{Relative maximum at }}\left( {\frac{8}{3},\frac{{16\sqrt 3 }}{9}} \right) \cr & {\text{No inflection points}} \cr & {\text{No horizontal asymptotes}} \cr} $$

Work Step by Step

$$\eqalign{ & y = x\sqrt {4 - x} \cr & 4 - x \geqslant 0,{\text{ then the domain of the function is: }}\left( { - \infty ,4} \right] \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & y = 0\sqrt {4 - 0} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}y = 0 \cr & 0 = x\sqrt {4 - x} \cr & x = 0{\text{ and }}\left( {4,0} \right) \cr & x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {4,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {x\sqrt {4 - x} } \right] \cr & y' = x\frac{d}{{dx}}\left[ {\sqrt {4 - x} } \right] + \sqrt {4 - x} \frac{d}{{dx}}\left[ x \right] \cr & y' = x\left( {\frac{{ - 1}}{{2\sqrt {4 - x} }}} \right) + \sqrt {4 - x} \left( 1 \right) \cr & y' = - \frac{x}{{2\sqrt {4 - x} }} + \sqrt {4 - x} \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & - \frac{x}{{2\sqrt {4 - x} }} + \sqrt {4 - x} = 0 \cr & \sqrt {4 - x} = \frac{x}{{2\sqrt {4 - x} }} \cr & 4 - x = \frac{x}{2} \cr & \frac{3}{2}x = 4 \cr & x = \frac{8}{3} \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ { - \frac{x}{{2\sqrt {4 - x} }} + \sqrt {4 - x} } \right] \cr & {\text{Using a CAS }}\left( {{\text{Wolfram Website}}} \right){\text{we obtain }} \cr & y'' = \frac{{3x - 16}}{{4{{\left( {4 - x} \right)}^{3/2}}}} \cr & {\text{Evaluate }}y''\left( x \right){\text{ at }}x = \frac{8}{3} \cr & y''\left( {\frac{8}{3}} \right) = \frac{{3\left( {8/3} \right) - 16}}{{4{{\left( {4 - 8/3} \right)}^{3/2}}}} < 0,{\text{ relative maximum at }}x = \frac{8}{3} \cr & f\left( {\frac{8}{3}} \right) = \frac{8}{3}\sqrt {4 - \frac{8}{3}} = \frac{{16\sqrt 3 }}{9} \to {\text{Relative maximum at }}\left( {\frac{8}{3},\frac{{16\sqrt 3 }}{9}} \right) \cr & {\text{Let }}y''\left( x \right) = 0 \cr & \frac{{3x - 16}}{{4{{\left( {4 - x} \right)}^{3/2}}}} = 0 \cr & 3x - 16 = 0 \cr & x = \frac{{16}}{3} \approx 5.33 \cr & \frac{{16}}{3}{\text{ It is not in the domain of }}f\left( x \right),{\text{ then}} \cr & {\text{There are no inflection points}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to - \infty } x\sqrt {4 - x} = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.