Problem 49 - Calculus: Early Transcendentals

Question

Which of the following functions \( f \) has a removable discontinuity at \( a \)? If the discontinuity is removable, find a function \( g \) that agrees with \( f \) for \( x \neq a \) and is continuous at \( a \). (a) \( f(x) = \dfrac{x^4 -1}{x - 1}\), \( a = 1 \) (b) \( f(x) = \dfrac{x^3 - x^2 - 2x}{x - 2} \), \( a = 2 \) (c) \( f(x) = [ \sin x ] \), \( a = \pi \)

Step-by-Step Solution

Answer

Functions (a) and (b) have removable discontinuities and can be made continuous by defining \( g(x) = (x + 1)(x^2 + 1) \) at \( x = 1 \) and \( g(x) = x(x + 1) \) at \( x = 2 \), respectively. Function (c) has a jump discontinuity at \( \pi \), which is not removable.

Step 1: Identify the Type of Discontinuity for (a)

For option (a), the function is \( f(x) = \frac{x^4 - 1}{x - 1} \). First, try to simplify the expression. Factor the numerator: \( x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1) \). Thus, \( f(x) = \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1} \). Cancel the common factor \( (x - 1) \), except at \( x = 1 \). So, \( f(x) = (x + 1)(x^2 + 1) \) for \( x eq 1 \). This indicates a removable discontinuity at \( x = 1 \), since the function can be redefined at this point.

Step 2: Find a Continuous Function for (a)

To find a function \( g \) that is continuous at \( a = 1 \) and agrees with \( f \) for \( x eq 1 \), define: \( g(x) = (x + 1)(x^2 + 1) \) for all \( x \), including \( x = 1 \). Evaluate \( g(1) = (1 + 1)(1^2 + 1) = 4 \). Hence, \( g(x) \) is continuous at \( x = 1 \).

Step 3: Identify the Type of Discontinuity for (b)

For option (b), the function is \( f(x) = \frac{x^3 - x^2 - 2x}{x - 2} \). Factor the numerator: \( x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x - 2)(x + 1) \). So, \( f(x) = \frac{x(x - 2)(x + 1)}{x - 2} \). Cancel the common factor \( (x - 2) \), except at \( x = 2 \). Thus, \( f(x) = x(x + 1) \) for \( x eq 2 \). This indicates a removable discontinuity at \( x = 2 \).

Step 4: Find a Continuous Function for (b)

To find a function \( g \) that is continuous at \( a = 2 \) and agrees with \( f \) for \( x eq 2 \), define: \( g(x) = x(x + 1) \) for all \( x \), including \( x = 2 \). Evaluate \( g(2) = 2(2 + 1) = 6 \). Hence, \( g(x) \) is continuous at \( x = 2 \).

Step 5: Check Continuity for (c)

For option (c), the function is \( f(x) = [ \sin x ] \), where \([ \cdot ]\) denotes the greatest integer function (floor function). This function evaluates to the greatest integer less than or equal to \( \sin x \). Since the value of \( f(x) \) jumps at integer values of \( \sin x \), there is a jump discontinuity at \( a = \pi \). Jump discontinuities are not removable.

Key Concepts

Continuity

Continuity of a function is a fundamental concept in calculus. It concerns whether a function exists without interruption over an interval. For a function to be continuous at a point, three conditions must be met:

• The function is defined at that point.
• The limit of the function as it approaches the point from both directions exists.
• The function’s value and the limit at that point are equal.

The concept of removable discontinuities is closely related. These occur when a function is not defined at a point or the limit is different from the function value, but we can "fix" the function by redefining it at that point. Thus, continuity is restored when you adjust a single point, often by simplifying the function and evaluating it at the limit point.

Function Simplification

Function simplification involves expressing a complex function in a simpler form. This technique is critical when identifying and addressing discontinuities, especially removable ones.

For example, consider the function \( f(x) = \frac{x^4 - 1}{x - 1} \). The numerator can be factored into \((x - 1)(x + 1)(x^2 + 1)\), allowing us to cancel the \( (x - 1) \) term when \( x eq 1 \). This simplification reveals that \( f(x) \) is really defined by \( (x + 1)(x^2 + 1) \) except at \( x = 1 \). Addressing the point of discontinuity by calculating the limit as \( x \) approaches 1 cultivates continuity. Simplification is a powerful tool for revealing the underlying behavior of functions.

Piecewise Function

A piecewise function is defined by multiple sub-functions, each applying to certain intervals of the domain. This setup is often used to manage different behaviors within a single function. When dealing with discontinuities, particularly removable ones, redefining a function piecewise can restore continuity.

For instance, if a function \( f \) is undefined or discontinuous at a point \( a \), we might redefine it by creating a new function \( g \), setting \( g(x) = f(x) \) where \( x eq a \) and assigning a value at \( x = a \) that aligns with the function’s limit. This technique ensures the function remains continuous across all points.

Greatest Integer Function

The greatest integer function, or the floor function, denoted by \([ \cdot ]\), outputs the greatest integer less than or equal to a given number. It produces a step-like graph with jump discontinuities at each integer value of its argument.

Considering a function like \( f(x) = [\sin x] \), the value jumps at each crossing of integer values, causing discontinuities that cannot be repaired by simple redefinition.

• Jump discontinuities like these cannot be considered removable.
• This is because the function inherently has steps with no way to smooth across these jumps.

The greatest integer function is a classic example of a non-continuous function that behaves in a discontinuous manner by design.