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 - Page 56: 1


a. $19$ b. $1$

Work Step by Step

a. $f(x)=x^3+1 , [2,3]$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$$ $f(x_{2})=f(3)=3^{3}+1$ $f(3)=28$ $f(x_{1})=f(2)=2^{3}+1$ $f(2)=9$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{f(3)-f(2)}{x_{2}-x_{1}}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{28-9}{3-2}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{19}{1}$$ $$\frac{\Delta{y}}{\Delta{x}}=19$$ b. $f(x)=x^3+1, [-1,1]$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$$ $f(x_{2})=f(1)=1^{3}+1$ $f(1)=2$ $f(x_{1})=f(-1)=-1^{3}+1$ $f(-1)=0$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{f(1)-f(-1)}{x_{2}-x_{1}}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{2-0}{1-(-1)}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{2}{2}$$ $$\frac{\Delta{y}}{\Delta{x}}=1$$
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