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Chapter 2: Problem 7

Find all \(x \in \mathbb{R}\) that satisfy the equation \(|x+1|+|x-2|=7\).

Short Answer

Expert verified
The solutions for the given equation are \( x = -5 \) and \( x = 4 \).

Step by step solution

01

Identify the Cases

Anytime you see absolute values in an equation, identify the unique cases that can make the expression inside the absolute value equal to zero, which would be \( x = -1 \) and \( x = 2 \) for this equation.
02

Break Down the Equation

Inject the identified cases into the equation to break down into three equations: \n\nCase 1: \( x < -1 \)\nIn this case both \( x + 1 \) and \( x - 2 \) are negative. Hence, the equation becomes \n-(x + 1) - (x - 2) = 7. Simplifying the equation gives x = -5.\n\nCase 2: \(-1 \leq x < 2\) \nHere, \( x + 1 \) is non-negative and \( x - 2 \) is negative. The equation becomes:\n (x + 1) - (x - 2) = 7. Simplifying the equation gives x = 4.\n\nCase 3: \( x \geq 2 \)\nIn this case, both \( x + 1 \) and \( x - 2 \) are non-negative. Hence, the equation becomes \n (x + 1) + (x - 2) = 7. Simplifying the equation gives x = 4.
03

Validate the Solutions

Check the calculated \( x \) values to ensure they fall within the ranges of \( x \) in their respective cases: \n\nFor Case 1: \( -5 < -1 \) is true. So, \( x = -5 \) is a solution.\n\nFor Case 2: \( -1 \leq 4 < 2 \) is false. So, \( x = 4 \) is not a solution in this case.\n\nFor Case 3: \( 4 \geq 2 \) is true. So, \( x = 4 \) is also a solution.

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Most popular questions from this chapter

Show that sup \(\\{1-1 / n: n \in \mathbb{N}\\}=1\)

Let \(S_{3}=\\{1 / n: n \in \mathbb{N}\\} .\) Show that sup \(S_{3}=1\) and inf \(S_{3} \geq 0 .\) (It will follow from the Archimedean Property in Section \(2.4\) that inf \(\left.S_{3}=0 .\right)\)

Let \(X\) be a nonempty set, and let \(f\) and \(g\) be defined on \(X\) and have bounded ranges in \(\mathbb{R}\). Show that $$ \sup \\{f(x)+g(x): x \in X\\} \leq \sup (f(x): x \in X\\}+\sup \\{g(x): x \in X\\} $$ and that $$ \inf \\{f(x): x \in X\\}+\inf \\{g(x): x \in X\\} \leq \inf \\{f(x)+g(x): x \in X\\} $$ Give examples to show that each of these inequalities can be either equalities or strict inequalities.

If \(a, b \in \mathbb{R}\), show that \(a^{2}+b^{2}=0\) if and only if \(a=0\) and \(b=0\).

Show that if \(a, b \in \mathbb{R}\) then (a) \(\max \\{a \cdot b\\}=\frac{1}{2}(a+b+|a-b|)\) and \(\min (a, b\\}=\frac{1}{2}(a+b-|a-b|)\). (b) \(\min \\{a, b, c\\}=\min \\{\min \\{a, b\\}, c)\).
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