Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.2 - The Derivative as a Function - Exercises 3.2 - Page 115: 18



Work Step by Step

Step 1. Given the function (replace $z$ with $x$ for convenience) $g(x)=1+\sqrt {4-x}$ and a point $(3,2)$, recall the formula for the differentiation of a function: $g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}$ Step 2. At $x=3$, $g'(3)=\lim_{h\to0}\frac{1+\sqrt {4-3-h}-1-\sqrt {4-3}}{h}=\lim_{h\to0}\frac{\sqrt {1-h}-1}{h}=\lim_{h\to0}\frac{(\sqrt {1-h}-1)(\sqrt {1-h}+1)}{h(\sqrt {1-h}+1)}=\lim_{h\to0}\frac{1-h-1}{h(\sqrt {1-h}+1)}=lim_{h\to0}\frac{-1}{\sqrt {1-h}+1}=-\frac{1}{2}$ which is the slope of the tangent line at point $(3,2)$ Step 3. The equation of the tangent line can be written as (use $z, w$): $w-2=-\frac{1}{2}(z-3)$ or $w=-\frac{1}{2}z+\frac{7}{2}$
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