Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.1 - Functions and Their Graphs - Exercises 1.1 - Page 12: 27

Answer

The first part of the graph is the parabolic function $4-x^2$ until the point (1,3), which cuts the x-axis at (-2,0). The second part of the graph is the parabolic function $x^2 +2x$ from the point (1,3) on through (2,8).
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Work Step by Step

$F(1) = 4 - 1^2$ $F(1) = 4 - 1$ $F(1) = 3$ The first part will cut the x-axis at: $4-x^2 = 0$ $x^2 = 4$ $x = \sqrt4$ or $x = -\sqrt4$ $x = 2$ or $x = -2$ So the first part of the graph will be the parabolic function $4-x^2$ until the point (1,3), which cuts the x-axis at (-2,0). $ F(1) = 1^2 + 2 \times 1$ $ F(1) = 1 + 2$ $ F(1) = 3$ $ F(2) = 2^2 + 2 \times 2$ $ F(2) = 4 + 4$ $ F(2) = 8$ So the second part of the graph will be the parabolic function $x^2 +2x$ from the point (1,3) on through (2,8).
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