Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 161: 55

Answer

\[\begin{align} & a.\text{ }x=-\frac{1}{2} \\ & b.\text{ the tangent line to the graph }f\left( x \right)=x{{e}^{2x}}\text{ } \\ & \text{at }x=-\frac{1}{2}\text{ is horizontal} \\ \end{align}\]

Work Step by Step

\[\begin{align} & \text{Let }f\left( x \right)=x{{e}^{2x}} \\ & \text{a}\text{.} \\ & \text{Calculate }f'\left( x \right) \\ & f'\left( x \right)=\frac{d}{dx}\left[ x{{e}^{2x}} \right] \\ & f'\left( x \right)=x\frac{d}{dx}\left[ {{e}^{2x}} \right]+{{e}^{2x}}\frac{d}{dx}\left[ x \right] \\ & f'\left( x \right)=x\left( 2{{e}^{2x}} \right)+{{e}^{2x}}\left( 1 \right) \\ & f'\left( x \right)=2x{{e}^{2x}}+{{e}^{2x}} \\ & \text{Let }f'\left( x \right)=0\text{ to find the values of }x\text{ at which the slope of} \\ & \text{the curve }f\left( x \right)\text{ is 0}\text{.} \\ & 2x{{e}^{2x}}+{{e}^{2x}}=0 \\ & {{e}^{2x}}\left( 2x+1 \right)=0 \\ & {{e}^{2x}}=0\text{ or }2x+1=0 \\ & {{e}^{2x}}\text{ is always }>\text{0 for all real number }x,\text{ then} \\ & 2x+1=0 \\ & x=-\frac{1}{2} \\ & \text{The value at which the derivative of the function is }0\text{ is }x=-\frac{1}{2}. \\ & \\ & \text{b}\text{. } \\ & \text{From the graph of }f\left( x \right)\text{ shown below we can say that the} \\ & \text{line tangent to the graph }f\left( x \right)=x{{e}^{2x}}\text{ at }x=-\frac{1}{2}\text{ is horizontal.} \\ & \text{Graph } \\ \end{align}\]
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.