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Chapter 2: Problem 11

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

Short Answer

Expert verified
The solution to the inequality is \( -\frac{1}{2} < x < 2 \).

Step by step solution

01

Determine The 4 Scenarios

Depending on whether \( x \) is less than or greater than -2 and 1, four scenarios arise: \n\n1) \( x < -2 \) and \( x < 1 \) \n2) \( x < -2 \) and \( x > 1 \) \n3) \( x > -2 \) and \( x < 1 \) \n4) \( x > -2 \) and \( x > 1 \) \n\nSince scenario 2 is not possible because \( x \) cannot be both less than -2 and greater than 1 at the same time, there are effectively three scenarios to check.
02

Check Scenario 1

For scenario 1 (\( x < -2 \) and \( x < 1 \)), both \( |x+2| \) and \( |x-1| \) become negative, which we can simplistically write as minus their original expressions. This yields the inequality \( 4 < -(x+2) - (x-1) < 5 \), or which simplifies to \( 4 < -2x + 3 < 5 \). Solving this, we find that there are no solutions for \( x \) that satisfy this condition.
03

Check Scenario 3

For scenario 3 (\( x > -2 \) and \( x < 1 \)), \( |x+2| \) becomes positive, and \( |x-1| \) becomes negative. The inequality thus becomes \( 4 < (x+2) - (x-1) < 5 \), or \( 4 < 2x + 3 < 5 \). Solving this, we find the interval valid for this scenario is \( -\frac{1}{2} < x < 1 \).
04

Check Scenario 4

For scenario 4 (\( x > -2 \) and \( x > 1 \)), both \( |x+2| \) and \( |x-1| \) are positive. The inequality is then simply \( 4 < (x+2) + (x-1) < 5 \), which simplifies to \( 4 < 2x + 1 < 5 \). Solving this, we find the interval valid for this scenario is \( 1 < x < 2 \).
05

Combine All Valid Intervals

By combining the valid intervals of scenarios 3 and 4, we find that all real numbers \( x \) that satisfy \( 4 < |x+2| + |x-1| < 5 \) are such that \( -\frac{1}{2} < x < 2 \).

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Most popular questions from this chapter

Give the two binary representations of \(\frac{3}{8}\) and \(\frac{7}{16}\).

Let \(S \subseteq \mathbb{R}\) be nonempty. Prove that if a number \(u\) in \(\mathbb{R}\) has the properties: (i) for every \(n \in \mathbb{N}\) the number \(u-1 / n\) is not an upper bound of \(S\), and (ii) for every number \(n \in \mathbb{N}\) the number \(u+1 / n\) is an upper bound of \(S\), then \(u=\sup S .\) (This is the converse of Exercise 2.3.8.)

Solve the following equations, justifying each step by referring to an appropriate property or theorem. (a) \(2 x+5=8\), (b) \(x^{2}=2 x\), (c) \(x^{2}-1=3\), (d) \((x-1)(x+2)=0\).

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

(a) If \(c>1\), show that \(c^{n} \geq c\) for all \(n \in \mathbb{N}\), and that \(c^{n}>c\) for \(n>1\). (b) If \(01\).
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