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Chapter 2: Problem 10

Find all \(x \in \mathbb{R}\) that satisfy the inequality \(4<|x+2|+|x-1|<5\).

Short Answer

Expert verified
The solution for the given inequality are all \(x \in (1.5, 2)\) and \(x \in (-1, -0.5)\).

Step by step solution

01

Analyze the Absolute Value Inequalities

The absolute value of a variable \(x\) can be either positive or negative. When inside the absolute value is greater than zero, we take it as positive. When inside the absolute value is less than zero, we take it as negative. We will separate both absolute values \(|x+2|\) and \(|x-1|\) into their respective cases.
02

Solving for \(x+2>0\) and \(x-1>0\)

Here, the inequality becomes \(4 < x+2 + x-1 < 5\). Simplifying the inequality leads to \(4 < 2x + 1 < 5\). Subtracting 1 from all parts gives \(3 < 2x < 4\). Finally, dividing each part of the inequality by 2 gives \(1.5 < x < 2\). So, when \(x > -2\) and \(x > 1\), the answer is \(x \in (1.5, 2)\).
03

Solving for \(x+2

Here, the inequality becomes \(4 < -(x+2) -(x-1) < 5\). Simplifying the inequality leads to \(4 < -2x +3 < 5\). Subtracting 3 from all parts gives \(1 < -2x < 2\). Dividing each part of the inequality by -2, and reversing the inequality signs, we get \(-0.5 > x > -1\). So, when \(x < -2\) and \(x < 1\), the answer is \(x \in (-1, -0.5)\).
04

Solving for \(x+2>0\) and \(x-1

Here, the inequality becomes \(4 < x+2 -(x-1) < 5\). Simplifying the inequality leads to \(4 < 3 < 5\), which is always true. So, there are no solutions when \(x>-2\) while \(x<1\).
05

Solving for \(x+20\)

Here, the inequality becomes \(4 < -(x+2) +(x-1) < 5\). Simplifying the inequality leads to \(4< -1 <5\), which is not true. So, there are no solutions when \(x<-2\) while \(x>1\).

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Most popular questions from this chapter

Let \(S_{2}=\\{x \in \mathbb{R}: x>0\\} .\) Does \(S_{2}\) have lower bounds? Does \(S_{2}\) have upper bounds? Does inf \(S_{2}\) exist? Does sup \(S_{2}\) exist? Prove your statements.

Find all \(x \in \mathbb{R}\) that satisfy the following inequalities: (a) \(|4 x-5| \leq 13\), (b) \(\left|x^{2}-1\right| \leq 3\).

(a) Show that if \(a>0\), then \(1 / a>0\) and \(1 /(1 / a)=a\). (b) Show that if \(a

If \(a, b \in \mathbb{R}\), show that \(|a+b|=|a|+|b|\) if and only if \(a b \geq 0\).

If \(0
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