Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 178: 69


$(\frac{5}{2},\frac{9}{4})$ and $(\frac{3}{2},-\frac{1}{4})$

Work Step by Step

Step 1. Find the derivative of the function $y=\frac{x}{2}+\frac{1}{2x-4}$ as $y'=\frac{1}{2}-\frac{1}{2(x-2)^2}$ Step 2. Let $y'=-3/2$, we have $\frac{1}{2}-\frac{1}{2(x-2)^2}=-3/2$ and $\frac{1}{2(x-2)^2}=2$ which gives $(x-2)^2=\frac{1}{4}$ and $x=2\pm\frac{1}{2}$ Step 3. We have $x_1=5/2$ and $x_2=3/2$, corresponding to $y_1=\frac{5/2}{2}+\frac{1}{2(5/2)-4}=\frac{5}{4}+1=\frac{9}{4}$ and $y_2=\frac{3/2}{2}+\frac{1}{2(3/2)-4}=\frac{3}{4}-1=-\frac{1}{4}$ Step 4. Thus the answer is yes and the points are $(\frac{5}{2},\frac{9}{4})$ and $(\frac{3}{2},-\frac{1}{4})$
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