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: 5

Answer

$y=2x+5$, see figure.
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Work Step by Step

Given the curve as $y=4-x^2$ and a point $(-1,3)$ on the curve, we can find the slope of the tangent line at this point as: $m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to0}\frac{4-(-1+h)^2-(4-(-1)^2)}{h}=\lim_{h\to0}\frac{2h-h^2}{h}=\lim_{h\to0}(2-h)=2$ The equation for the tangent line at $(-1,3)$ can then be written as $y-3=2(x+1)$ or $y=2x+5$, as shown in the figure.
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