Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 1 - Functions and Their Graphs - Section 1.3 Properties of Functions - 1.3 Assess Your Understanding - Page 76: 64

Answer

(a). $-2$, (b). $0$, (c). $5$.

Work Step by Step

(a). From $x_1=-1$ to $x_2=1$, we have the average rate of change as $\frac{h(x_2)-h(x_1)}{x_2-x_1}=\frac{((1)^2-2(1)+3)-((-1)^2-2(-1)+3)}{1+1}=-2$, (b). From $x_1=0$ to $x_2=2$, we have the average rate of change as $\frac{h(x_2)-h(x_1)}{x_2-x_1}=\frac{((2)^2-2(2)+3)-((0)^2-2(0)+3)}{2-0}=0$, (c). From $x_1=2$ to $x_2=5$, we have the average rate of change as $\frac{h(x_2)-h(x_1)}{x_2-x_1}=\frac{((5)^2-2(5)+3)-((2)^2-2(2)+3)}{5-2}=5$.
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