Problem 76 - College Algebra with Corequisite Support

Question

For the following exercises, identify the removable discontinuity. $$ f(x)=\frac{x^{3}+1}{x+1} $$

Step-by-Step Solution

Answer

The removable discontinuity is at \( x = -1 \).

Step 1: Identify Discontinuity

The function \( f(x) = \frac{x^3 + 1}{x + 1} \) is potentially discontinuous wherever the denominator is zero. So, set the denominator equal to zero and solve: \( x + 1 = 0 \), giving \( x = -1 \). This indicates a discontinuity at \( x = -1 \).

Step 2: Simplify the Function

To determine if the discontinuity at \( x = -1 \) is removable, we simplify the function by factoring the numerator.The numerator \( x^3 + 1 \) is a sum of cubes, which can be factored as: \( (x + 1)(x^2 - x + 1) \).The function simplifies to: \( f(x) = \frac{(x + 1)(x^2 - x + 1)}{x + 1} \).

Step 3: Cancel Common Factors

Observe that the \( (x + 1) \) terms in the numerator and denominator cancel each other out:\[ f(x) = x^2 - x + 1 \] when \( x eq -1 \).

Step 4: Determine the Removable Discontinuity

The common factor \( x + 1 \) being canceled indicates a removable discontinuity.Thus, there is a removable discontinuity at \( x = -1 \).The simplified function without discontinuity is \( f(x) = x^2 - x + 1 \).

Step 5: Define the Function Continuously

The function \( f(x) = \frac{x^3 + 1}{x + 1} \) can be rewritten to remove the discontinuity by setting it equal to the simplified form at \( x = -1 \):\[ f(x) = \begin{cases} x^2 - x + 1, & \text{if} \ x eq -1 \ 1, & \text{if} \ x = -1 \end{cases} \]This provides a continuous extension of \( f(x) \) over all real numbers.

Key Concepts

Factoring Polynomials

Polynomials are fundamental to understanding complex mathematical expressions, and factoring them is an essential skill. In our original exercise, we encounter the polynomial expression in the numerator: \(x^3 + 1\). This is an example of a sum of cubes. It can be factored into \((x + 1)(x^2 - x + 1)\). This is an application of a general formula for factoring the sum of cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).
The ability to factor polynomials allows us to simplify expressions and solve equations effectively. By factoring \(x^3 + 1\), we reveal a common factor with the denominator. This common factor is critical to identifying points of discontinuity in the rational function we are analyzing. Moreover, factoring simplifies the computation of complex algebraic equations.
When you encounter a polynomial that needs simplifying, try to recognize patterns like the sum or difference of cubes. Applying these formulas correctly will often lead to simplifications that make solving functions much easier.

Rational Functions

Rational functions are ratios of two polynomials. In this scenario, the function \(f(x) = \frac{x^3 + 1}{x + 1}\) represents a typical rational function. These types of functions have potential discontinuities, which occur where the denominator equals zero and the function becomes undefined.
By setting \(x + 1 = 0\), we find a discontinuity at \(x = -1\). This is where our initial rational function encountered an issue. However, after factoring and canceling the common term \(x + 1\), we transition to \(f(x) = x^2 - x + 1\) for all \(x eq -1\). This elimination of the discontinuity demonstrates how simplifying a rational function by factoring helps us understand its behavior across its domain.
Understanding rational functions and their properties, such as discontinuities, is essential for graphing and analyzing these functions. They can represent real-world situations where variables are related in a proportional manner.

Continuous Functions

Continuous functions have no breaks, jumps, or holes across their domain. The concept of continuity is crucial in calculus and real-world applications where smooth operations are necessary.
In our derived function \(f(x) = x^2 - x + 1\), after addressing the removable discontinuity, we turned our initial rational expression into a function that is continuous for all real numbers except at the point of discontinuity. By defining \(f(x)\) to explicitly handle the point \(x = -1\), as in \[ f(x) = \begin{cases} x^2 - x + 1, & \text{if} \ x eq -1\ 1, & \text{if} \ x = -1 \end{cases} \] we have ensured the function is continuous everywhere.
In general, continuous functions are advantageous because they can be modeled mathematically without any unexpected breaks. They are easy to integrate and differentiate, further establishing their importance in calculus and beyond.