Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1174: 32

Answer

a. See graph and explanations. b. $y=\frac{1}{2}x+2$ c. see graph.
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Work Step by Step

a. Given $f(x)=\sqrt {x-1}+2$, we can obtain the graph of $f(x)$ by the transformation of $g(x)=\sqrt x$: shift horizontally 1 unit to the right and then 2 units up as shown in the figure. b. At $x=2$, we have $f(2)=\sqrt {2-1}+2=3$, and the point is $(2,3)$. The slope of the tangent can be found as $m=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{\sqrt {2+h-1}+2-\sqrt {2-1}-2}{h}=\lim_{h\to0}\frac{\sqrt {1+h}-1}{h}=\lim_{h\to0}\frac{1+h-1}{h(\sqrt {1+h}+1)}=\lim_{h\to0}\frac{1}{\sqrt {1+h}+1}=\frac{1}{2}$ Thus, the slope-intercept equation of the tangent line is $y-3=\frac{1}{2}(x-2)$ or $y=\frac{1}{2}x+2$ c. see graph.
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