Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises: 67

Answer

\[\bar c{\,^,}\left( x \right) = \frac{{ - x - 2}}{{2{x^2}\sqrt {x + 1} }}\]

Work Step by Step

\[\begin{gathered} c\,\left( x \right) = \sqrt {x + 1} \hfill \\ The\,\,average\,\,cost\,\,is\,\,defined\,\,by \hfill \\ \bar c\,\left( x \right) = \frac{{c\,\left( x \right)}}{x} \hfill \\ \bar c\,\left( x \right) = \frac{{\sqrt {x + 1} }}{x} \hfill \\ The\,\,marginal\,\,cost\,\,is\,\,given\,\,by \hfill \\ \frac{d}{{dx}}\,\,\left[ {\bar c\,\left( x \right)} \right] = \frac{d}{{dx}}\,\,{\left[ {\frac{{\sqrt {x + 1} }}{x}} \right]^,} \hfill \\ By\,\,the\,\,quotient\,\,rule \hfill \\ \bar c{\,^,}\left( x \right) = \frac{{x\,{{\left( {\sqrt {x + 1} } \right)}^,} - \,\left( {\sqrt {x + 1} } \right)\,{{\left( x \right)}^,}}}{{{x^2}}} \hfill \\ \bar c{\,^,}\left( x \right) = \frac{{x\,\left( {\frac{1}{{2\sqrt {x + 1} }}} \right) - \,\left( {\sqrt {x + 1} } \right)}}{{{x^2}}} \hfill \\ Simplifying \hfill \\ \bar c{\,^,}\left( x \right) = \frac{{x - 2\,{{\left( {\sqrt {x + 1} } \right)}^2}}}{{2{x^2}\sqrt {x + 1} }} \hfill \\ \bar c{\,^,}\left( x \right) = \frac{{ - x - 2}}{{2{x^2}\sqrt {x + 1} }} \hfill \\ \hfill \\ \end{gathered} \]
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