Question

Describe the largest set \(S\) on which it is correct to say that \(f\) is continuous. \(f(x, y)=\frac{x^{2}+x y-5}{x^{2}+y^{2}+1}\)

Short answer

Function \( f(x, y) \) is continuous for all real \((x, y) \in \mathbb{R}^2\).

Step-by-step solution

Understanding the Function

The function given is \( f(x, y)=\frac{x^{2}+x y-5}{x^{2}+y^{2}+1} \). This is a two-variable rational function since it is the ratio of polynomials.

Rational Functions and Continuity

A rational function is continuous everywhere it is defined. It is undefined where the denominator is equal to zero. Therefore, \( f(x, y) \) is undefined when \( x^{2}+y^{2}+1 = 0 \).

Investigating the Denominator

Let's find the values of \( x \) and \( y \) for which the denominator \( x^{2}+y^{2}+1 \) equals zero. Solving \( x^{2} + y^{2} + 1 = 0 \), we see this equation has no real solutions, because \( x^{2} + y^{2} \) are always non-negative and \( 1 \) is positive.

Conclusion about the Domain

Since the denominator is never zero for any real numbers \( x \) and \( y \), \( f(x, y) \) is defined and continuous for all real \( x \) and \( y \). Therefore, the largest set \( S \) on which \( f \) is continuous is the set of all real numbers \( \mathbb{R}^2 \).

Key concepts

Rational Functions

Rational functions are an essential concept in mathematics. They are expressions that represent a ratio of two polynomials. An example is a function of the form \( f(x, y) = \frac{P(x, y)}{Q(x, y)} \), where \( P(x, y) \) and \( Q(x, y) \) are polynomials and \( Q(x, y) eq 0 \). Rational functions can have variables in both their numerator and denominator.

A rational function is continuous where it is defined. The points where the denominator becomes zero are excluded from the domain, since division by zero is undefined. In these cases, the function has a discontinuity.

Consider the rational function given in the problem: \( f(x, y) = \frac{x^2 + xy - 5}{x^2 + y^2 + 1} \). Here, \( x^2 + xy - 5 \) is the numerator, and \( x^2 + y^2 + 1 \) is the denominator. The task is to ensure that the denominator does not equal zero, to find where the function is continuous.

Domain of a Function

The domain of a function refers to all the possible input values (or points in the case of two-variable functions) for which the function is defined. Determining the domain involves identifying any restrictions on the input values.

In rational functions like \( f(x, y) = \frac{x^2 + xy - 5}{x^2 + y^2 + 1} \), the domain is determined by the denominator. For continuity, the denominator should not be zero. Therefore, to find the domain, we first solve the equation \( x^2 + y^2 + 1 = 0 \). However, since \( x^2 \) and \( y^2 \) are always non-negative, and \( 1 \) is positive, this equation has no real solutions. Thus, the domain of this function is all pairs \((x, y)\) in \( \mathbb{R}^2 \), indicating the function is defined everywhere for real \( x \) and \( y \).

This also means the function is continuous across its domain, which covers the entire xy-plane.

Two-Variable Functions

Two-variable functions, such as \( f(x, y) \), depend on two different input variables. This leads to a three-dimensional graph where \( f(x, y) \) represents a surface in \( \mathbb{R}^3 \).

The continuity of such a function depends on the function being defined for every possible pair of \( x \) and \( y \) within its domain. For \( f(x, y) = \frac{x^2 + xy - 5}{x^2 + y^2 + 1} \), the function is continuous over its entire domain since we established that the denominator \( x^2 + y^2 + 1 \) is never zero.

Continuity in two-variable functions ensures that small changes in \( x \) or \( y \) result in small changes in the function value, a vital property for applications like modeling natural phenomena and solving real-world problems involving several inputs. Two-variable functions offer a rich ground for exploring how variables interact on a plane, providing insights that are often crucial for engineering, physics, and economics.