Question

Continuity of a Function In Exercises \(27-30,\) discuss the continuity of each function. $$ f(x)=\frac{x^{2}-1}{x+1} $$

Short answer

The function \(f(x) = \frac{x^2 - 1}{x+1}\) is continuous for all \(x \neq -1\). The function has a removable discontinuity at \(x = -1\).

Step-by-step solution

Simplify the Function

First, the given function should be simplified if possible. Notice that the denominator of the function is not defined at \(x = -1\). This indicates that there might be a discontinuity at \(x = -1\). We can try to simplify the numerator to see if it is possible to cancel out these possible undefined points. The numerator \(x^2-1\) is a difference of squares and can be factored into \((x-1)(x+1)\). So, the function \(f(x)\) can be rewritten as \(f(x) = \frac{(x-1)(x+1)}{x+1}\).

Cancel out Common Factors

In this step, cancel the common factors in the numerator and the denominator. The term \(x+1\) can be cancelled from the numerator and the denominator of the function. Therefore, the function becomes \(f(x) = x - 1\).

Analyze the Continuity of the Simplified Function

Now, the simplified function \(f(x) = x - 1\) is a linear function, which is continuous for all values of \(x\). The original function \(f(x) = \frac{x^2 - 1}{x+1}\) is therefore continuous for all \(x \neq -1\). However, at \(x = -1\), the original function is not defined (the denominator becomes zero), leading to a removable discontinuity.

Confirm the Discontinuity

To confirm the discontinuity at \(x = -1\), we use limits. If the limit of the function \(f(x)\) as \(x\) approaches -1 is equal to the value of the function at \(x = -1\), then the function is continuous at that point. We know that the function is not defined at \(x = -1\), so we try to find the limit of the function as \(x\) approaches -1. Taking the limit of the simplified function \(f(x) = x - 1\) as \(x\) approaches -1, we get -2. Thus, the function is not continuous at \(x = -1\), as suspected.

Key concepts

Removable Discontinuity

A discontinuity in a function can be perplexing, but understanding removable discontinuity is straightforward. Consider a function that is not defined at a specific point yet can be "fixed" or "removable" by modifying its definition slightly at that point, such that it becomes continuous.

In the exercise, the function \( f(x) = \frac{x^2 - 1}{x+1} \) shows a discontinuity at \( x = -1 \). This is because the denominator becomes zero, making the function undefined there. However, upon simplifying the function to \( f(x) = x - 1 \), the expression is continuous everywhere except where we have the original problem.

• To "remove" the discontinuity, we can redefine the function explicitly at \( x = -1 \) to match the limit of the rest of the function.
• A removable discontinuity does not affect the overall shape or direction of the function's graph.
• Mathematically, this type of discontinuity can often be addressed through algebraic simplification or substitution.

Recognizing and handling removable discontinuities is essential when ensuring functions behave well over their entire domain.

Factoring Polynomials

Factoring polynomials is a critical skill in algebra and calculus, often used to simplify expressions and solve equations. A polynomial can frequently be broken down, or "factored," into the product of two or more simpler polynomials, making them easier to manage.

In the given exercise, the function \( f(x) = \frac{x^2 - 1}{x+1} \) involves a numerator that can be factored as a difference of squares: \( x^2 - 1 = (x - 1)(x + 1) \).

• The factoring process identifies common factors that exist in both the numerator and the denominator, allowing these to be canceled, simplifying the function.
• This simplification reveals the true nature of the function and provides insights into its domain and possible discontinuities.
• Factoring is not just essential for simplification but is also vital in finding roots and analyzing polynomial expressions.

Mastery of factoring polynomials greatly aids in identifying limits, solving quadratic equations, and simplifying rational expressions like the one in the exercise.

Limits in Calculus

Limits are foundational to calculus and indispensable for understanding function continuity and behavior at specific points. The concept of a limit helps assess what value a function approaches as the input approaches a particular point.

In the exercise solution, analyzing the limit of \( f(x) = x - 1 \) as \( x \) approaches \(-1\) highlights why the original function has a discontinuity at that point.

• The limit constructs an expectation of where the function value should reach, helping to confirm its continuity or discontinuity.
• By evaluating \( \lim_{x \to -1} (x - 1) = -2 \), we confirmed the discontinuity since the real value of the function could not be defined at \( x = -1 \).
• Limits can also be evaluated using different techniques like substitution, factoring (as seen here), and even L'Hôpital's rule for more complicated indeterminate forms.

A solid grasp of limits opens the door to deeper exploration of derivatives, integrals, and the overall topology of function behavior in calculus.