Problem 67 - Brief Applied Calculus

Question

Determine whether each function is continuous or discontinuous. If discontinuous, state where it is discontinuous. $$ f(x)=\frac{x+1}{x-1} $$

Step-by-Step Solution

Answer

The function is discontinuous at \( x = 1 \).

Step 1: Define Continuity

A function is continuous at a point if the limit of the function as it approaches the point from both directions equals the function's value at that point. Mathematically, a function \( f \) is continuous at \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \). A function is continuous on its domain if it is continuous at every point within the domain.

Step 2: Identify Potential Points of Discontinuity

The function \( f(x) = \frac{x+1}{x-1} \) is a rational function, which might be discontinuous where the denominator is zero. Thus, set the denominator equal to zero to find potential discontinuities: \( x-1 = 0 \). Solving this gives \( x = 1 \).

Step 3: Analyze the Continuity at \( x = 1 \)

At \( x = 1 \), the denominator of \( f(x) \) becomes zero, making the function undefined at that point. Therefore, \( f(x) \) does not have a value at \( x = 1 \), and thus cannot be continuous there.

Step 4: Determine the Type of Discontinuity

Since \( f(x) \) is not defined at \( x = 1 \), there is a point discontinuity (also known as a removable discontinuity) at \( x = 1 \). The function is otherwise continuous where it is defined.

Key Concepts

Rational Functions

A rational function is a type of function that can be represented as the ratio of two polynomials. In mathematical terms, it looks like \( f(x) = \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). This formula means that the function is defined for all values of \( x \), except for those where the denominator \( Q(x) \) equals zero.

When dealing with rational functions, you always need to check the denominator for any values of \( x \) that might cause it to become zero. These values are where the rational function might be discontinuous. For instance, in the function \( f(x) = \frac{x+1}{x-1} \), the denominator \( x-1 \) becomes zero when \( x = 1 \). Hence, this specific function is not defined at \( x = 1 \), and evaluating the function at this point would result in division by zero, which is undefined.

Understanding rational functions is crucial in calculus as they illustrate key concepts about how algebraic expressions behave and where they might fail to be continuous.

Limits and Continuity

Limits are a foundational concept in calculus that help us understand the behavior of functions as they approach a particular point. In simple terms, the limit of a function \( f(x) \) as \( x \) approaches some value \( a \), is the value that \( f(x) \) gets close to as \( x \) gets closer and closer to \( a \). Continuity uses this idea of limits to define whether a function is continuous at a given point.

A function is considered continuous at a point \( x = a \) if three conditions are met:

• The function \( f(x) \) is defined at \( x = a \).
• The limit of \( f(x) \) as \( x \) approaches \( a \) exists.
• The limit of \( f(x) \) equals the function value \( f(a) \).

If any one of these conditions fails, then the function is discontinuous at that particular point. In the example of \( f(x) = \frac{x+1}{x-1} \), the function is discontinuous at \( x = 1 \) because the function is not defined there, resulting in the absence of a limit that equals a function value.

Removable Discontinuity

A removable discontinuity occurs in a function at a certain point where the function is not defined or does not have a value. These discontinuities are named 'removable' because, under some conditions, you could "remove" the discontinuity by appropriately defining the function at the point so that it becomes continuous.

This usually happens when there is a factor in both the numerator and the denominator that cancels out. For example, with the function \( g(x) = \frac{(x+1)(x-2)}{x-2} \), the term \( x-2 \) cancels, leaving \( g(x) = x+1 \) for all points except \( x=2 \). While \( x = 2 \) makes the original expression undefined, plugging it back into \( g(x) = x + 1 \) gives a value, thus "removing" the discontinuity.
In the case of \( f(x) = \frac{x+1}{x-1} \), at \( x = 1 \), the function is undefined, showcasing a typical removable discontinuity, yet it's not always possible to simply redefine it as the structure doesn't allow it to cancel out terms.