Problem 530 - Precalculus

Question

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

Step-by-Step Solution

Answer

Removable discontinuity at \(x=-1\).

Step 1: Find the point of discontinuity

To find the points of discontinuity, note that the denominator is zero when \(x=-1\). This suggests that there might be a discontinuity at this point.

Step 2: Check if the point is removable

To determine if the discontinuity at \(x=-1\) is removable, simplify the function if possible. Factor the numerator \(x^3+1\) as follows:\[x^3+1 = (x+1)(x^2-x+1)\]The function now becomes:\[\frac{(x+1)(x^2-x+1)}{x+1}\]Since the factor \(x+1\) cancels out with the denominator, the function simplifies to \(x^2-x+1\) when \(x eq -1\).

Key Concepts

Rational Functions

Rational functions are mathematical expressions that represent the ratio of two polynomials. They take the form \( f(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \) is not zero. These functions can model a wide range of real-world situations and are often used in algebra and calculus.

Understanding rational functions involves learning about the behavior of these functions when the denominator approaches zero, as this is where discontinuities may occur. Discontinuities are points in a function where it is not defined or not continuous. For rational functions, discontinuities often occur at the roots of the denominator.

To fully grasp rational functions, it's helpful to consider:

• Domains: The set of all possible input values for which the function is defined, usually all real numbers except where the denominator is zero.
• Zeros: Values of \( x \) for which \( f(x) = 0 \), found by setting the numerator equal to zero.
• Vertical asymptotes: Lines \( x = a \) where the function approaches infinity, often at points where the denominator is zero and non-removable.

By examining these aspects, one can predict the behavior of rational functions and deal with discontinuities effectively.

Factoring Polynomials

Factoring polynomials is a critical skill in algebra. It involves breaking down a polynomial into simpler "factor" polynomials that, when multiplied together, yield the original polynomial.

Consider the polynomial \( x^3 + 1 \) from the original exercise. To factor this, we use sum of cubes, as \( x^3 + 1 = (x+1)(x^2-x+1) \). This method is useful for simplifying expressions and solving for roots effectively.

When dealing with polynomial expressions, factoring can help in:

• Simplification: Reduce complex expressions to simpler ones which are easier to handle and understand.
• Finding zeros: Since a product is zero if any factor is zero, it helps find where the polynomial equals zero.
• Simplifying rational functions: As seen in this exercise, factoring can help identify and remove common factors in the numerator and denominator.

Mastering polynomial factoring is not only useful for solving equations but also necessary for understanding more complex algebraic expressions, like rational functions.

Discontinuity in Functions

Discontinuities in functions are points where a function is not continuous, meaning the function's value does not approach a fixed number as \( x \) approaches from either side. In rational functions, discontinuities generally occur where the denominator is zero.

A discontinuity can be categorized as removable or non-removable.

• Removable discontinuity: Occurs when factors in the numerator and denominator cancel out, like in \( \frac{(x+1)(x^2-x+1)}{x+1} \) where \( x+1 \) cancels. This implies the function could be made continuous elsewhere by redefining it.
• Non-removable discontinuity: Typically occurs with vertical asymptotes, where the function value goes to infinity.

Understanding discontinuities is crucial in calculus for analyzing functions' behavior and ensuring mathematical models are accurate and continuous wherever intended. By factoring polynomials and simplifying expressions, one can identify and often "remove" potential discontinuities, making the function easier to interpret and graph.