Problem 13 Determine and sketch the set of ... [FREE SOLUTION]

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Introduction to Real Analysis

Robert G. Bartle, Donald R. Sherbert

$Math Studyset 91影视 Explanations$ Math

3 Edition

Chapter 2: Problem 13

Determine and sketch the set of pairs $(x, y)$ in $\mathbb{R} \times \mathbb{R}$ that satisfy: (a) $|x| \leq|y|$, (b) $|x|+|y| \leq 1$, (c) $|x y| \leq 2$, (d) $|x|-|y| \geq 2$.

Short Answer

Expert verified

For the given inequalities, graphing them results in several regions in the ($x$, $y$) plane. The presented inequalities represent various geometric regions, including diamond shapes (|x| + |y| 鈮� 1) and overlapping half-planes (|x| 鈮� |y|). Additionally, the pairs that satisfy |xy| 鈮� 2 are located within a total of four regions, determined by the intersecting hyperbolas y = 卤2/x. Finally the inequality |x| - |y| 鈮� 2 result in sketch of four half-planes obtained.

Step by step solution

Graphing $|x| \leq|y|$

This inequality exhibit values where $y$ is either greater than or equal to both $x$ and $-x$, thus essentially gets divided into two half planes. One is $y \geq x$ (1st and 2nd quadrants) and $y \geq -x$ (2nd and 3rd quadrants). In the coordinate system, draw two perpendicular lines representing these equations and shade the area of their overlap.

Graphing $|x|+|y| \leq 1$

This is the equation of a diamond centered at the origin with vertices at $(1,0), (-1,0), (0,1), (0,-1)$. Each of these constituent inequalities represents a half plane and the overlap is a diamond, therefore simply plot these points and draw a diamond shape around them and shade the inside area.

Graphing $|x y| \leq 2$

This one is a bit more complex. It represents four regions in a plane bounded by hyperbolas $y = 2/x$ and $y = -2/x$. When $x > 0, y \geq -2/x \text{ and } y \leq 2/x$ and when $x < 0, y \leq -2/x \text{ and } y \geq 2/x$. These are the two graph branches in the 1st and 3rd quadrants. Similar logic applies to the 2nd and 4th quadrants.

Graphing $|x|-|y| \geq 2$

This inequality, when analyzed results in four separate cases. Case 1: When $x \geq 0$ and $y \geq 0$, it simplifies to $x - y \geq 2$. Case 2: When $x \geq 0$ and $y < 0$, it simplifies to $x + y \geq 2$. Case 3: When $x < 0$ and $y \geq 0$, it simplifies to $-x - y \geq 2$. Case 4: When $x < 0$ and $y < 0$, it simplifies to $-x + y \geq 2$. Each of these cases describes a half-plane, and the total graph is the union of these four half-planes.

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91影视

Determine and sketch the set of pairs \((x, y)\) in \(\mathbb{R} \times \mathbb{R}\) that satisfy: (a) \(|x| \leq|y|\), (b) \(|x|+|y| \leq 1\), (c) \(|x y| \leq 2\), (d) \(|x|-|y| \geq 2\).

Short Answer

Step by step solution

Graphing \(|x| \leq|y|\)

Graphing \(|x|+|y| \leq 1\)

Graphing \(|x y| \leq 2\)

Graphing \(|x|-|y| \geq 2\)

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