University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.1 - Rates of Change and Tangents to Curves - Exercises: 2

Answer

a. $2$ b. $0$

Work Step by Step

a. $g(x)=x^2-2x , [1,3]$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{g(x_{2})-g(x_{1})}{x_{2}-x_{1}}$$ $g(x_{2})=g(3)=3^{2}-2(3)$ $g(3)=3$ $g(x_{1})=g(1)=1^{2}-2(1)$ $g(1)=-1$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{g(3)-g(1)}{x_{2}-x_{1}}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{3-(-1)}{3-1}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{4}{2}$$ $$\frac{\Delta{y}}{\Delta{x}}=2$$ b. $g(x)=x^2-2x , [-2,4]$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{g(x_{2})-g(x_{1})}{x_{2}-x_{1}}$$ $g(x_{2})=g(4)=4^{2}-2(4)$ $g(4)=8$ $g(x_{1})=g(-2)=(-2)^{2}-2(-2)$ $g(-2)=8$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{g(4)-g(-2)}{x_{2}-x_{1}}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{8-8}{4-(-2)}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{0}{6}$$ $$\frac{\Delta{y}}{\Delta{x}}=0$$
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