University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 117: 5

Answer

The equation of the tangent line is $y=2x+5$.
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Work Step by Step

The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$y=f(x)=4-x^2\hspace{1cm}A(-1,3)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$m=\lim_{h\to0}\frac{4-(a+h)^2-(4-a^2)}{h}=\lim_{h\to0}\frac{-(a+h)^2+a^2}{h}$$ $$m=\lim_{h\to0}\frac{-a^2-2ah-h^2+a^2}{h}=\lim_{h\to0}\frac{-2ah-h^2}{h}$$ $$m=\lim_{h\to0}(-2a-h)=-2a-0=-2a$$ $$m=-2\times(-1)=2$$ 2) Find the equation of the tangent line at $A(-1,3)$: The tangent line would have this form: $$y=2x+m$$ Substitute $A(-1,3)$ here to find $m$: $$2\times(-1)+m=3$$ $$-2+m=3$$ $$m=5$$ So the equation of the tangent line at $A$ is $y=2x+5$.
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