Answer
See graph, $y=\sqrt[3] {x-1},\ x\ge1$
Work Step by Step
1. See graph for $(t^{3/2}+1,\sqrt t)$ over $t\ge0$
2. From $y=\sqrt t$, we have $t^{1/2}=y$, thus $x=y^3+1$ or $y=\sqrt[3] {x-1},\ x\ge1$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 9 - Analytic Geometry - Section 9.7 Plane Curves and Parametric Equations - 9.7 Assess Your Understanding - Page 714: 18
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