Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.8 - Inverse Functions - Exercise Set - Page 272: 102

Answer

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1569142195

Work Step by Step

First, simplify the above expression. Then divide the above expression by 3: $\frac{3\left| 2x-1 \right|}{3}\ge \frac{21}{3}$ Further, the above expression becomes $\left| 2x-1 \right|\ge 7$ So the absolute value or modulus function is denoted by $\left| x \right|.$ Here, x is a real number and the absolute value or modulus function of x is defined as $\left| x \right|=\left\{ \begin{align} & \,\,\,x,\,x\ge 0 \\ & -x,\,x<0 \\ \end{align} \right.$ So, $\left| 2x-1 \right|=\left( 2x-1 \right)$ or $\left| 2x-1 \right|=-\left( 2x-1 \right)$ As $\left| 2x-1 \right|\ge 7,$ $\left( 2x-1 \right)\ge 7$ or $-\left( 2x-1 \right)\ge 7$ Now solve the above inequality: $\left( 2x-1 \right)\ge 7$ Add 1 to both sides of the above inequality: $2x\ge 8$ Divide both sides of the above inequality by 2: So, $x\ge 4.$ Now, solve the inequality $-\left( 2x-1 \right)\ge 7$: $-2x+1\ge 7$ Subtract 1 from both sides of the above equation: $-2x\ge 6$ Divide both sides of the above inequality by −2: When the inequality is multiplied or divided by a negative number, the sign of inequality gets reversed. So, $x\le -3.$ The solution set is $\left\{ x|-3\ge x\ge 4 \right\}$ of the inequality $3\left| 2x-1 \right|\ge 21$.
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