Rational Function

A rational function is a quotient of polynomial functions and plays an important role in calculus. Understanding the continuity properties of rational functions is essential for studying more complicated types of function.

Basic Properties

A rational function has the following basic properties:

  • Domain: all real numbers except the points where the denominator is zero
  • Continuity: continuous on its domain
  • Points of discontinuity: discontinuous at the points where the denominator is zero
  • Asymptotes: may have a vertical asymptote and a horizontal asymptote

Definition of Rational Functions

Definition of a Rational Function

A function of the form f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)} is called a rational function, where P(x)P(x) and Q(x)Q(x) are both polynomial functions and Q(x)≠0Q(x) \neq 0.

Common Rational Functions

1. Simple Rational Functions

Example 1: f(x)=1xf(x) = \frac{1}{x}

Analysis:

  • Domain: x≠0x \neq 0
  • Discontinuous at x=0x = 0
  • Has a vertical asymptote x=0x = 0
  • Has a horizontal asymptote y=0y = 0

Example 2: f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}

Analysis:

  • Domain: x≠1x \neq 1
  • Discontinuous at x=1x = 1
  • Can be simplified to f(x)=x+1f(x) = x + 1 (when x≠1x \neq 1)
  • If we define f(1)=2f(1) = 2, then the function is continuous at x=1x = 1

2. Complicated Rational Functions

Example 3: f(x)=1x2+1f(x) = \frac{1}{x^2 + 1}

Analysis:

  • Domain: R\mathbb{R} (the denominator is always greater than zero)
  • Continuous everywhere on R\mathbb{R}
  • The graph is a smooth curve
  • Has a horizontal asymptote y=0y = 0

Example 4: f(x)=x2+1xf(x) = \frac{x^2 + 1}{x}

Analysis:

  • Domain: x≠0x \neq 0
  • Discontinuous at x=0x = 0
  • Has a vertical asymptote x=0x = 0
  • Has no horizontal asymptote

Continuity Analysis of Rational Functions

Determining Continuity

The continuity of a rational function f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)}:

  1. Continuous on its domain: continuous at every point where Q(x)≠0Q(x) \neq 0
  2. Points of discontinuity: discontinuous at every point where Q(x)=0Q(x) = 0
  3. Removable discontinuities: if P(x0)=Q(x0)=0P(x_0) = Q(x_0) = 0 and lim⁡x→x0P(x)Q(x)\lim_{x \to x_0} \frac{P(x)}{Q(x)} exists, then the function can be made continuous at that point by defining f(x0)f(x_0)

Types of Point of Discontinuity

  1. Removable discontinuity: the limit exists but the function value is undefined
  2. Jump discontinuity: the left and right limits exist but are not equal
  3. Infinite discontinuity: the limit tends to infinity

Graph Characteristics of Rational Functions

Summary of Graph Features

  1. Smoothness: on its domain, the graph of a rational function is a smooth curve
  2. Asymptotes: may have a vertical asymptote and a horizontal asymptote
  3. Points of discontinuity: at the points where the denominator is zero there is a vertical asymptote or a removable discontinuity
  4. Local behaviour: near a point of continuity, the function values change gradually

Applications of Rational Functions

1. Function Modelling

Rational functions are widely used in science and engineering:

  • Physical modelling: describe various physical phenomena
  • Economic modelling: describe economic relationships such as costs and revenues
  • Engineering applications: describe various engineering parameters

2. Function Approximation

Rational functions can be used to approximate other complicated functions:

  • Padé approximation: approximate a complicated function with rational functions
  • Interpolation: construct a rational function through given points

Exercises

Exercise 1

Determine the continuity of f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2} at x=2x = 2.

rational functioncontinuityfunction continuity

Approach: Check whether the function is defined at that point.

Steps:

  1. When x=2x = 2, the denominator x−2=0x - 2 = 0
  2. Therefore the function is undefined at x=2x = 2
  3. The function is discontinuous at x=2x = 2

Answer: The function is discontinuous at x=2x = 2, because the function is undefined at that point.

Exercise 2

Determine the continuity of f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1} at x=1x = 1.

rational functioncontinuityfunction continuity

Approach: Analyse the domain and the continuity of the rational function.

Steps:

  1. When the denominator is zero: x−1=0x - 1 = 0, that is, x=1x = 1
  2. Domain: x≠1x \neq 1, that is, (−∞,1)∪(1,+∞)(-\infty, 1) \cup (1, +\infty)
  3. A rational function is continuous on its domain
  4. Therefore the interval of continuity is (−∞,1)∪(1,+∞)(-\infty, 1) \cup (1, +\infty)

Answer: The function is discontinuous at x=1x = 1, and the interval of continuity is (−∞,1)∪(1,+∞)(-\infty, 1) \cup (1, +\infty).

Exercise 3

Let f(x)={x2−1x−1,x≠1a,x=1f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ a, & x = 1 \end{cases}. Find the value of the constant aa that makes f(x)f(x) continuous at x=1x = 1.

rational functioncontinuityfunction continuity

Approach: Use the definition of continuity so that the function value equals the limit value.

Steps:

  1. Compute the limit: lim⁡x→1x2−1x−1=lim⁡x→1(x−1)(x+1)x−1=lim⁡x→1(x+1)=2\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2
  2. By the definition of continuity: f(1)=lim⁡x→1f(x)f(1) = \lim_{x \to 1} f(x)
  3. Therefore a=2a = 2

Answer: a=2a = 2.

Exercise 4

Let f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}. Find the intervals of continuity of f(x)f(x).

rational functioncontinuityfunction continuity

Approach: Analyse the domain and the continuity of the rational function.

Steps:

  1. When the denominator is zero: x−1=0x - 1 = 0, that is, x=1x = 1
  2. Domain: x≠1x \neq 1, that is, (−∞,1)∪(1,+∞)(-\infty, 1) \cup (1, +\infty)
  3. A rational function is continuous on its domain
  4. Therefore the interval of continuity is (−∞,1)∪(1,+∞)(-\infty, 1) \cup (1, +\infty)

Answer: The interval of continuity is (−∞,1)∪(1,+∞)(-\infty, 1) \cup (1, +\infty).

Exercise 5

Determine the continuity of f(x)=x3+1x2−1f(x) = \frac{x^3 + 1}{x^2 - 1} at x=1x = 1.

rational functioncontinuityfunction continuity

Approach: Check whether the function is defined at that point.

Steps:

  1. When x=1x = 1, the denominator x2−1=0x^2 - 1 = 0
  2. Therefore the function is undefined at x=1x = 1
  3. The function is discontinuous at x=1x = 1

Answer: The function is discontinuous at x=1x = 1, because the function is undefined at that point.

Exercise 6

Let f(x)={x3−1x−1,x≠1a,x=1f(x) = \begin{cases} \frac{x^3 - 1}{x - 1}, & x \neq 1 \\ a, & x = 1 \end{cases}. Find the value of the constant aa that makes f(x)f(x) continuous at x=1x = 1.

rational functioncontinuityfunction continuity

Approach: Use the definition of continuity so that the function value equals the limit value.

Steps:

  1. Compute the limit: lim⁡x→1x3−1x−1=lim⁡x→1(x−1)(x2+x+1)x−1=lim⁡x→1(x2+x+1)=3\lim_{x \to 1} \frac{x^3 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{x-1} = \lim_{x \to 1} (x^2+x+1) = 3
  2. By the definition of continuity: f(1)=lim⁡x→1f(x)f(1) = \lim_{x \to 1} f(x)
  3. Therefore a=3a = 3

Answer: a=3a = 3.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
R\mathbb{R}Mathematical symbolBlackboard bold R (Real numbers)The set of all real numbers

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