/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 (a) find the function's domain, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 27

(a) find the function's domain, (b) find the function's range, (c) describe the function's level curves, (d) find the boundary of the function's domain, (e) determine if the domain is an open region, a closed region, or neither, and ( \(\mathbf{f}\) ) decide if the domain is bounded or unbounded. $$f(x, y)=\sin ^{-1}(y-x)$$

Short Answer

Expert verified
The domain is \( -1 \leq y - x \leq 1 \), range is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), level curves are lines \( y = x + \sin(c) \), boundary: lines \( y = x \pm 1 \), domain is closed and unbounded.

Step by step solution

01

Finding the Function's Domain

The domain of the function is the set of all points \((x, y)\) for which the function is defined. Since \(f(x, y) = \sin^{-1}(y - x)\), the argument \(y - x\) must satisfy \(-1 \leq y - x \leq 1\) for \(\sin^{-1}\) to be defined. Therefore, the domain consists of all points \((x, y)\) such that \(-1 \leq y - x \leq 1\). This represents the region between the lines \(y = x - 1\) and \(y = x + 1\).
02

Finding the Function's Range

The range of the function is the set of all possible values of \(f(x, y)\). For \(f(x, y) = \sin^{-1}(y - x)\), as \(y - x\) varies from \(-1\) to \(1\), the function value \(f\) varies from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). Hence, the range of the function is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
03

Describing the Function's Level Curves

Level curves are given by the equation \(\sin^{-1}(y - x) = c\), where \(c\) is a constant. This rearranges to \(y - x = \sin(c)\). Therefore, each level curve is a line given by \(y = x + \sin(c)\), where \(c\) is a constant in \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
04

Finding the Boundary of the Function's Domain

The boundary of the domain is where \(y - x = -1\) or \(y - x = 1\). This means the domain is bounded by the lines \(y = x - 1\) and \(y = x + 1\). Thus, the boundary consists of these two lines.
05

Determining if the Domain is Open, Closed, or Neither

An open region does not include its boundary; a closed region does include its boundary. By the definition \(-1 \leq y - x \leq 1\), the lines \(y = x - 1\) and \(y = x + 1\) are part of the domain. Therefore, the domain includes its boundary and is a closed region.
06

Deciding if the Domain is Bounded or Unbounded

A domain is bounded if it is contained within some finite region of the plane. The domain, defined by \(-1 \leq y - x \leq 1\), is a strip of infinite length extending along the x-axis. Thus, the domain is unbounded, as it extends infinitely in both positive and negative directions along the \(x\)-axis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function's Domain
The domain of a function refers to all the possible input values for which the function is defined. For the function \(f(x, y) = \sin^{-1}(y - x)\), the domain is dictated by the requirement that the argument of \(\sin^{-1}\) must lie within the range \([-1, 1]\). This translates to the inequality \(-1 \leq y - x \leq 1\). Breaking this down:
  • The function is defined wherever \(y - x\) falls between \(-1\) and \(1\).
  • Geometrically, this represents a strip or band running parallel to the line \(y = x\), but shifted between the lines \(y = x - 1\) and \(y = x + 1\).
Hence, this region forms the complete set of inputs where the function is properly defined.
Level Curves
Level curves are an incredible way to visualize functions of two variables. For \(f(x, y) = \sin^{-1}(y - x)\), each level curve represents all points that yield the same function value \(c\). Analytically, it is given by the equation \(\sin^{-1}(y - x) = c\), which simplifies to \(y - x = \sin(c)\).
  • A level curve, therefore, represents a line defined by \(y = x + \sin(c)\).
  • Each line parallel to \(y = x\) shifts based on \(c\), where \(c\) itself is restricted to \([\frac{\pi}{2}, -\frac{\pi}{2}]\), ensuring the result of \(\sin(c)\) always lies within \([-1, 1]\).
This visualization aids in understanding how the function's output changes across the plane.
Bounded and Unbounded Regions
Identifying whether the domain of a function is bounded or unbounded is crucial in understanding its behavior. Given \(-1 \leq y - x \leq 1\) for \(f(x, y) = \sin^{-1}(y - x)\), this forms a region extending infinitely along the \(x\)-axis. Here's what this all means:
  • The domain is described as a strip equidistant from the line \(y = x\), specified by boundaries \(y = x - 1\) and \(y = x + 1\).
  • While the strip has finite height due to its boundaries, it stretches indefinitely along the \(x\)-axis.
This setup renders the domain an unbounded region because there is no finite limit to its extent horizontally.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: $$f(x, y)=x y$$ Curves: i) The line \(x=2 t, \quad y=t+1\) ii) The line segment \(x=2 t, \quad y=t+1, \quad-1 \leq t \leq 0\) iii) The line segment \(x=2 t, \quad y=t+1, \quad 0 \leq t \leq 1\)

You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer.

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at the critical point \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field of the planet is weakest. The planet is spherical, with a radius of 6 units. Based on a coordinate system whose origin is at the center of the planet, the strength of the magnetic field is given by \(M(x, y, z)=6 x-\) \(y^{2}+x z+60 .\) Where should you locate the radio telescope?

For what values of the constant \(k\) does the Second Derivative Test guarantee that \(f(x, y)=x^{2}+k x y+y^{2}\) will have a saddle point at (0,0)\(?\) A local minimum at (0,0)\(?\) For what values of \(k\) is the Second Derivative Test inconclusive? Give reasons for your answers.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.