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At what points of \(\mathbb{R}^{2}\) is a rational function of two variables continuous?

Short Answer

Expert verified
Answer: A rational function of two variables is continuous at all points in 鈩澛 except for the points where its denominator is equal to zero.

Step by step solution

01

Define a rational function of two variables

Let \(f(x, y)\) be a rational function of two variables \(x\) and \(y\). We can represent it as follows: $$ f(x, y) = \frac{p(x, y)}{q(x, y)} $$ where \(p(x, y)\) and \(q(x, y)\) are polynomials in variables \(x\) and \(y\).
02

Identify the condition for continuity

A rational function is continuous wherever its denominator is nonzero. Therefore, we must find the points in \(\mathbb{R}^{2}\) where the denominator \(q(x, y)\) is not equal to zero: $$ q(x, y) \neq 0 $$
03

Find the solution set

The rational function \(f(x, y)\) will be continuous at all points \((x, y)\) in \(\mathbb{R}^{2}\) that satisfy the condition \(q(x, y) \neq 0\). The points where the rational function is not continuous are the points where the denominator \(q(x, y)\) is equal to zero: $$ q(x, y) = 0 $$ To find the points in \(\mathbb{R}^{2}\) where the rational function is continuous, we need to exclude the points where the denominator \(q(x,y)\) is equal to zero from the entire \(\mathbb{R}^{2}\). To conclude, a rational function of two variables is continuous at all points in \(\mathbb{R}^{2}\) except for the points where its denominator is equal to zero. The continuity of a rational function depends on its denominator, so it is not possible to give a general answer for all rational functions. However, it can be determined for any specific rational function by analyzing its denominator.

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Key Concepts

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

Rational Functions
Rational functions are intriguing, especially when dealing with several variables such as in multivariable calculus. A rational function with two variables typically looks like this:\[f(x, y) = \frac{p(x, y)}{q(x, y)}\]Here, \(p(x, y)\) and \(q(x, y)\) are polynomials involving the variables \(x\) and \(y\). Think of a rational function as a fraction where the numerator and denominator are both polynomial expressions. Rational functions can be both useful and tricky because they may not be defined everywhere. Their domain is generally the largest set of points where the denominator is nonzero, implying potential challenges with continuity when dividing by zero. In many applications, checking the behavior of rational functions is crucial, particularly at points where the denominator approaches zero. Understanding their domain and assessments for continuity can help prevent errors in calculations, ultimately clarifying many functional behaviors.
Polynomials
Polynomials are the building blocks of rational functions, and they have a very straightforward structure. A polynomial in two variables \(x\) and \(y\) might be expressed as:\[p(x, y) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 + b_my^m + b_{m-1}y^{m-1} + ... + b_1y + b_0\]These expressions are composed of terms with coefficients (like \(a_n\) and \(b_m\)) multiplied by powers of the variables. They exhibit a smooth, continuous nature throughout their domain, making them easier to handle than general functions.When crafting a rational function, the numerator and denominator are typically polynomials. The continuity and smoother behavior of polynomials play a significant role in determining where the overall rational function remains continuous. Analyzing the structure of these polynomial components aids in finding where the rational function might face issues, especially if the denominator becomes zero.
Denominator Nonzero Condition
The concept of a denominator being nonzero is pivotal in the context of rational functions because it dictates where these functions are continuous. For a rational function \(f(x, y) = \frac{p(x, y)}{q(x, y)}\), continuity is guaranteed at all points where \(q(x, y)\) does not equal zero.Let's break it down:
  • Continuity: A function is continuous at a point if the value of the function at that point equals the limit of the function as it approaches that point from any direction.
  • Denominator's Role: If \(q(x, y) = 0\), the function is undefined at that point, leading to a discontinuity.
  • Finding Points of Continuity: Evaluate and solve the equation \(q(x, y) = 0\) to understand where discontinuities might occur. Often, this involves algebraic manipulation or graphing.
For instance, if \(q(x, y) = y - x^2\), then the rational function is undefined along the line \(y = x^2\). Recognizing where \(q(x, y)\) hits zero allows you to map areas on the function's domain without continuous behavior, enabling better grasp and management of the rational function's properties.

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Most popular questions from this chapter

When two electrical resistors with resistance \(R_{1}>0\) and \(R_{2}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}.\) a. Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega\) and \(R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega\) b. Is it true that if \(R_{1}=R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases, then \(R\) is approximately unchanged? Explain. c. Is it true that if \(R_{1}\) and \(R_{2}\) increase, then \(R\) increases? Explain. d. Suppose \(R_{1}>R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases. Does \(R\) increase or decrease?

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Let the equation of the best-fit line be \(y=m x+b,\) where the slope \(m\) and the \(y\) -intercept \(b\) must be determined using the least squares condition. First assume that there are three data points \((1,2),(3,5),\) and \((4,6) .\) Show that the function of \(m\) and \(b\) that gives the sum of the squares of the vertical distances between the line and the three data points is $$ \begin{aligned} E(m, b)=&((m+b)-2)^{2}+((3 m+b)-5)^{2} \\ &+((4 m+b)-6)^{2} \end{aligned}. $$ Find the critical points of \(E\) and find the values of \(m\) and \(b\) that minimize \(E\). Graph the three data points and the best-fit line.

Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=2 x^{2}+y^{2}+2 x-3 y ; R=\left\\{(x, y): x^{2}+y^{2} \leq 1\right\\}$$

An identity Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Power functions and percent change Suppose that \(z=f(x, y)=x^{a} y^{b},\) where \(a\) and \(b\) are real numbers. Let \(d x / x, d y / y,\) and \(d z / z\) be the approximate relative (percent) changes in \(x, y,\) and \(z,\) respectively. Show that \(d z / z=a(d x) / x+b(d y) / y ;\) that is, the relative changes are additive when weighted by the exponents \(a\) and \(b.\)

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