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 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 647: 80

Answer

$ (2,\frac{13}{5}]$

Work Step by Step

1. Based on the given conditions, we have: $\frac{3}{x-2}\ge5\Longrightarrow \frac{3}{x-2}-5\ge0\Longrightarrow \frac{13-5x}{x-2}\ge0$ 2. Boundary points $x=2,\frac{13}{5}$, form intervals $(-\infty,2),(2,\frac{13}{5}],[\frac{13}{5},\infty)$ 3. Choose test values $x=0,2.5,3$, test the inequality and get results $False,\ True,\ False$ 4. Thus, we have the solution interval: $ (2,\frac{13}{5}]$
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