Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 174: 48


$$L(x)\approx -8x +48$$

Work Step by Step

Given $$ g(x)=\frac{x^{2}}{x-3}, \quad a=4$$ Since $g(4)=16$ and \begin{align*} g'(x)&=\frac{(x-3) 2 x-x^{2}(1 )}{(x-3)^{2}}\\ &=\frac{2 x^{2}-6 x-x^{2}}{(x-3)^{2}}\\ g'(4)&=-8 \end{align*} Then the linearization at $a=4$ is given by \begin{align*} L(x)&\approx g^{\prime}(a)(x-a)+f(a)\\ &=-8\left(x-4\right)+16\\ &=-8x +48 \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.