/*! 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 40 So the input (3,5) yields an out... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

So the input (3,5) yields an output of \(\sqrt{2} .\) We define the do- main for this function just as we did in Chapter 3: The domain is the set of all inputs that yield real-number outputs. For instance, the ordered pair (1,4) is not in the domain of the function we have been discussing, because (as you should check for yourself\() f(1,4)=\sqrt{-1},\) which is not a real number. We can determine the domain of the function in equation ( 1 ) by requiring that the quantity under the radical sign be non negative. Thus we require that \(2 x-y+1 \geq 0\) and, consequently, \(y \leq 2 x+1\) (Check this.) The following figure shows the graph of this inequality; the domain of our function is the set of ordered pairs making up the graph. In Exercises follow a similar procedure and sketch the domain of the given function. (Graph cant copy) $$h(x, y)=\ln (x y)$$

Short Answer

Expert verified
The domain consists of points in the first and third quadrants where \( xy > 0 \).

Step by step solution

01

Understand the function

We are given the function \( h(x, y) = \ln(xy) \). This function involves the natural logarithm \( \ln \). For \( \ln(a) \) to be defined and return a real number, it is required that \( a > 0 \).
02

Determine the condition for the domain

Since \( h(x, y) = \ln(xy) \), we need the argument \( xy \) to be greater than 0, i.e., \( xy > 0 \). This condition arises because the logarithm of a non-positive number is undefined in the real number system.
03

Express the inequality in terms of x and y

The inequality \( xy > 0 \) implies that either both \( x \) and \( y \) are positive, or both are negative. This creates two separate conditions: \( x > 0 \) and \( y > 0 \), or \( x < 0 \) and \( y < 0 \).
04

Sketch the domain based on the inequality

The domain of the function \( h(x, y) = \ln(xy) \) in the coordinate plane is composed of two regions: one in the first quadrant where both \( x \) and \( y \) are positive, and one in the third quadrant where both \( x \) and \( y \) are negative. There are no intersections in the second or fourth quadrants since those would require one of the variables to be positive and the other to be negative, which would make \( xy \leq 0 \).

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 91Ó°ÊÓ!

Key Concepts

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

Real Numbers
Real numbers are the building blocks of mathematics that we use in our everyday life. They include all the numbers we usually work with:
  • Positive and negative numbers, like -3 and 7
  • Whole numbers and fractions, such as 0, 4.5, and -2/3
  • Irational numbers, such as \(\pi\) and \(\sqrt{2}\)
Real numbers can be found on the number line, stretching infinitely in both the positive and negative directions.
This concept is crucial when determining the domain of a function, as our function will typically need to output real numbers. Domains that output non-real numbers, such as complex numbers, are often excluded when we're dealing with functions involving real numbers.
Inequalities
Inequalities are mathematical expressions used to compare two values or expressions and state any of the following:
  • Greater than ( > )
  • Less than ( < )
  • Greater than or equal to ( \(\geq\) )
  • Less than or equal to ( \(\leq\) )
In our example, we encounter the inequality \(xy > 0\). It indicates that both \(x\) and \(y\) are either positive or negative at the same time.
The inequality marks crucial boundaries, helping us discern where the function has a defined, real-valued output.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. In mathematics, the natural logarithm is key to continuous growth processes found in scientific fields.
For the natural logarithm to return a real number, the input must be a positive real number. This condition emerges directly from the domain properties of the logarithm function: it cannot handle non-positive numbers, effectively making \( \ln(xy) \) valid only when \( xy > 0 \).
This requirement shapes the domain of the function \( h(x, y) = \ln(xy) \). It helps ensure that operations involving natural logarithms return real numbers.
Coordinate Plane
The coordinate plane is a two-dimensional surface defined by two axes: the horizontal axis, known as the x-axis, and the vertical axis, known as the y-axis. Each point on this plane is represented by a pair of numbers \((x, y)\).
This plane is divided into four quadrants:
  • The first quadrant (both \(x\) and \(y\) are positive)
  • The second quadrant (\(x\) is negative, \(y\) is positive)
  • The third quadrant (both \(x\) and \(y\) are negative)
  • The fourth quadrant (\(x\) is positive, \(y\) is negative)
In our example, the domain of the function is comprised of two regions in the coordinate plane. It resides in the first and third quadrants, where the condition \(xy > 0\) holds true, delineating an area where both \(x\) and \(y\) share the same sign.

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

A curve \(y=x^{3}+A x^{2}+B x+C\) passes through the three points \((1,-2),(2,3),\) and (3,20) (a) Write down a system of three linear equations satisfied by \(A, B,\) and \(C\) (b) Solve the system in part (a). (c) With the values for \(A, B,\) and \(C\) that you found in part (b), use a graphing utility to draw the curve \(y=x^{3}+A x^{2}+B x+C,\) checking to see that it appears to pass through the three given points.

Find an example of two \(2 \times 2\) matrices \(A\) and \(B\) for which \(A B=0\) but neither \(A\) nor \(B\) is 0.

Let \(A=\left(\begin{array}{rrr}1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5\end{array}\right)\) (a) Compute the matrix product \(A A .\) What do you observe? (b) Use the result in part (a) to solve the following system. $$ \left\\{\begin{aligned} x-6 y+3 z &=19 / 2 \\ 2 x-7 y+3 z &=11 \\ 4 x-12 y+5 z &=19 \end{aligned}\right. $$

A manufacturer of office chairs makes three models: Utility, Secretarial, and Managerial. Three materials common to the manufacturing process for all of the models are cloth, steel, and plastic. The amounts of these materials required for one chair in each category are specified in the following table. The company wants to use up its inventory of these materials because of upcoming design changes. How many of each model should the manufacturer build to deplete its current inventory consisting of 476 units of cloth, 440 units of steel, and 826 units of plastic? $$\begin{array}{lccc} & \text { Utility } & \text { Secretarial } & \text { Managerial } \\ \hline \text { Cloth } & 3 & 4 & 2 \\ \text { Steel } & 2 & 5 & 8 \\ \text { Plastic } & 6 & 4 & 1 \\ \hline \end{array}$$

Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right) .\) Let \(A^{2}\) and \(B^{2}\) denote the matrix products \(A A\) and \(B B\), respectively. Compute each of the following. (a) \((A+B)(A+B)\) (b) \(A^{2}+2 A B+B^{2}\) (c) \(A^{2}+A B+B A+B^{2}\)

See all solutions

Recommended explanations on Math Textbooks

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.