Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Problems Plus - Problems - Page 274: 27

Answer

$k = 2\sqrt{e}$

Work Step by Step

If $e^{2x} = k~\sqrt{x}$ has one solution, then the curves $y = e^{2x}$ and $y = k~\sqrt{x}$ have one point of intersection and the two curves are tangent to each other at this point. If the two curves intersect at one point: $e^{2x} = k~\sqrt{x}$ $k = \frac{e^{2x}}{\sqrt{x}}$ If the two curves are tangent to each other at one point, then the derivatives are equal at this point: $2e^{2x} = \frac{k}{2\sqrt{x}}$ $k = 4~\sqrt{x}~e^{2x}$ We can equate the two expressions for $k$ to find the value of $x$ where the curves are tangent to each other: $k = 4~\sqrt{x}~e^{2x} = \frac{e^{2x}}{\sqrt{x}}$ $x = \frac{1}{4}$ We can find the value of $k$: $e^{2x} = k~\sqrt{x}$ $e^{2(\frac{1}{4})} = k~\sqrt{\frac{1}{4}}$ $k = 2e^{1/2}$ $k = 2\sqrt{e}$
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