Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 108: 18


$m=\frac{1}{6}$, $y-3= \frac{1}{6}(x-8)$

Work Step by Step

$f(x)=\sqrt {x+1}$ $f'(x)=\frac{d}{dx}(\sqrt {x+1})=\frac{1}{2\sqrt {x+1}}$ $m=f'(8)=\frac{1}{2\sqrt {8+1}}=\frac{1}{6}$ Equation of a straight line passing through a given point $(x_{0},y_{0})$ having finite slope m is given by $y-y_{0}=m(x-x_{0})$ Therefore, equation of the tangent at (8,3) to the curve $f(x)= \sqrt {x+1}$ is given by $y-3= \frac{1}{6}(x-8)$
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