University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 205: 152



Work Step by Step

The linear approximation of $f(x)$ at $x=0$ is given as: $L(x)=f(0)+x f'(0)$ We are given that $f(x)=\dfrac{2}{1-x}+\sqrt {1-x}-3.1$ and $f(0)=-0.1$ Also, $f'(x)=\dfrac{1}{(1-x)^2}+\dfrac{1}{2 \sqrt {1+x}}$ and $f'(0)=2.5$ Hence, $L(x)=(-0.1)+(x)(2.5) =-0.1+2.5x$
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.