Problem 23 - Thomas Calculus

Question

At what points are the functions in Exercises 13-30 continuous? $$y=\frac{x \tan x}{x^{2}+1}$$

Step-by-Step Solution

Answer

The function is continuous everywhere except at \( x = \frac{(2k+1)\pi}{2} \), where \( k \) is an integer.

Step 1: Definition of Continuity

A function is continuous at a point if it is defined at that point, and its limit as it approaches that point equals its value at that point. That is, for a function \( f(x) \) to be continuous at \( x = a \), the following must hold: \( \lim_{x \to a} f(x) = f(a) \).

Step 2: Function Analysis

Given the function \( y = \frac{x \tan x}{x^2 + 1} \), this function is a rational function, which means it is generally continuous except where the denominator is zero. However, in this case, the denominator \( x^2 + 1 \) is never zero for real numbers. Thus, potential points of discontinuity must come from the \( \tan x \) function.

Step 3: Trigonometric Function Considerations

The tangent function \( \tan x \) is discontinuous at points where it is undefined. Specifically, \( \tan x \) is not defined at \( x = \frac{(2k+1)\pi}{2} \), where \( k \) is any integer, because the tangent becomes infinite (vertical asymptotes occur).

Step 4: Conclusion of Continuity

The given function \( y = \frac{x \tan x}{x^2 + 1} \) is continuous everywhere except at points where \( \tan x \) is undefined. Therefore, \( y \) is continuous for all \( x \) except at \( x = \frac{(2k+1)\pi}{2} \), where \( k \) is any integer.

Key Concepts

Trigonometric Functions

Trigonometric functions are a core part of mathematics, especially important in analyzing periodic phenomena. One such function is the tangent function, denoted as \( \tan x \). It is derived from sine and cosine functions with the relation \( \tan x = \frac{\sin x}{\cos x} \). Because of this ratio, the tangent function exhibits unique behavior:

• It is periodic with a period of \( \pi \).
• It has vertical asymptotes where the cosine function equals zero, as dividing by zero is undefined.

These vertical asymptotes occur at \( x = \frac{(2k+1)\pi}{2} \), where \( k \) is any integer. This means the function goes to positive or negative infinity, creating discontinuities. Understanding these points helps us determine where functions involving \( \tan x \) are continuous.

Rational Functions

Rational functions are defined as the quotient of two polynomials. In general, a rational function \( f(x) = \frac{p(x)}{q(x)} \) is continuous everywhere except where the denominator \( q(x) \) equals zero. At these points, the function becomes undefined.In the exercise example, the rational component is \( \frac{x \tan x}{x^2 + 1} \). Here:

• The numerator is \( x \tan x \), a product of linear and trigonometric expressions.
• The denominator is \( x^2 + 1 \), which is polynomial.
• The denominator does not become zero for any real number \( x \), indicating it is continuous as \( x^2 + 1 > 0 \) for all real \( x \).

As a result, the continuity of our function does not depend on the rational expression's polynomial numerator or denominator directly but rather on the internal trigonometric component, \( \tan x \).

Points of Discontinuity

Points of discontinuity in a function are values of \( x \) where the function is not continuous. For rational functions containing trigonometric components like \( \frac{x \tan x}{x^2 + 1} \), these discontinuities often arise from the trigonometric part.As mentioned, the points where \( \tan x \) is undefined determine the discontinuities in this function. These occur at:

• \( x = \frac{(2k+1)\pi}{2} \) for any integer \( k \). This results in vertical asymptotes.

At these points, evaluating the function leads to infinity or undefined behavior. Outside these specific values, the function is well-behaved and continuous, integrating both the polynomial and trigonometric aspects harmoniously. Recognizing these points allows for accurate analysis and understanding of the function's behavior across its domain.