Problem 30 - Calculus Early Transcendentals

Question

\(29-38\) Determine the set of points at which the function is continuous. $$F(x, y)=\frac{x-y}{1+x^{2}+y^{2}}$$

Step-by-Step Solution

Answer

The function is continuous for all \( (x, y) \) in the real plane.

Step 1: Identify the Function Type

The function given is \( F(x, y) = \frac{x-y}{1+x^2+y^2} \). It is a rational function because it is expressed as a fraction of two polynomial functions.

Step 2: Determine Points of Discontinuity

In a rational function \( \frac{p(x, y)}{q(x, y)} \), the points of discontinuity occur where the denominator \( q(x, y) = 0 \). Here, \( q(x, y) = 1 + x^2 + y^2 \). Determine when \( 1 + x^2 + y^2 = 0 \).

Step 3: Analyze the Denominator

The equation \( 1 + x^2 + y^2 = 0 \) has no real solutions because \( x^2 + y^2 \geq 0 \) for all real \( x \) and \( y \), and adding 1 makes it always greater than zero. Therefore, \( q(x, y) eq 0 \) for any real \( x \) and \( y \).

Step 4: Conclude Continuity

Since the denominator never equals zero for real numbers, the function \( F(x, y) \) is continuous for all points in the real plane. There are no discontinuities in the domain of real numbers.

Key Concepts

Rational Functions

A rational function is a kind of mathematical expression that represents the ratio of two polynomials. A polynomial is a mathematical expression consisting of variables, coefficients, and constants that are combined using addition, subtraction, multiplication, and non-negative integer exponents. The general form of a rational function is:

• \( F(x, y) = \frac{P(x, y)}{Q(x, y)} \)

where \( P(x, y) \) and \( Q(x, y) \) are both polynomials. For the function to be considered rational, the denominator \( Q(x, y) \) must not be zero, as dividing by zero is undefined in mathematics. This rule leads naturally to the discussion of points of discontinuity, which occur where the denominator is zero. Understanding these conditions is key to exploring the continuity and domain of rational functions.

Points of Discontinuity

Points of discontinuity in functions are places where the function does not become continuous. For rational functions, these points typically occur where the denominator equals zero, rendering the function undefined at those specific coordinates. Given the rational function

• \( F(x, y) = \frac{x-y}{1+x^2+y^2} \)

we need to examine the conditions under which the denominator

• \( 1 + x^2 + y^2 \)

becomes zero to find points of discontinuity. However, since \( x^2 \) and \( y^2 \) are non-negative for all real \( x \) and \( y \), adding one ensures this expression is strictly positive. As a result, in the context of real numbers, no such points of discontinuity exist for this function. The function is continuous across its entire domain, which is the real plane.

Real Plane

The real plane, often referred to as the Cartesian plane, is a fundamental concept in mathematical analysis representing a two-dimensional space composed of all possible ordered pairs \( (x, y) \), where both \( x \) and \( y \) are real numbers. Features of the real plane include:

• It is infinite and covers every real number.
• Each point corresponds to a specific \( x \)-coordinate and \( y \)-coordinate.
• Functions defined on the real plane evaluate to a real number for each valid \( (x, y) \) coordinate pair.

For the function \( F(x, y) = \frac{x-y}{1+x^2+y^2} \), the real plane forms the function's domain. This continuity over the real plane implies the function behaves predictably everywhere in this space, without any holes or breaks.

Denominator Analysis

Analyzing the denominator of a rational function gives significant insight into where the function might be undefined. For

• \( F(x, y) = \frac{x-y}{1+x^2+y^2} \)

our denominator is

• \( 1 + x^2 + y^2 \)

The goal is to determine when this expression could be zero. However, because both \( x^2 \) and \( y^2 \) are non-negative, the smallest value \( x^2 + y^2 \) can obtain is zero, occurring only when both \( x \) and \( y \) are zero. Adding 1, however, shifts this sum away from zero, ensuring that the expression is always positive and never zero. This straightforward analysis confirms that there are no denominatorial issues across the real plane, leading to the conclusion that the function is continuous everywhere in this mathematical space.