Problem 79 - Brief Applied Calculus

Question

By canceling the common factor, \(\frac{(x-1)(x+2)}{x-1}\) simplifies to \(x+2\). At \(x=1\), however, the function \(\frac{(x-1)(x+2)}{x-1}\) is discontinuous (since it is undefined where the denominator is zero), whereas \(x+2\) is continuous. Are these two functions, one obtained from the other by simplification, equal to each other? Explain.

Step-by-Step Solution

Answer

The two functions are not equal because they differ at \(x=1\), where the original function is undefined.

Step 1: Identify the Expression

The given function is \( \frac{(x-1)(x+2)}{x-1} \). To simplify this, we will look for a common factor in the numerator and the denominator.

Step 2: Simplify the Expression

Notice that both the numerator \((x-1)(x+2)\) and the denominator \(x-1\) have a common factor of \(x-1\). We can cancel this common factor to simplify the expression to \(x+2\). However, this simplification holds only for values of \(x\) where \(x-1\) is not zero.

Step 3: Evaluate Discontinuity

The simplification process removes the factor that causes the function to be undefined at \(x=1\). Originally, \(\frac{(x-1)(x+2)}{x-1}\) is undefined for \(x=1\) because division by zero is not permissible. Thus, a hole in the graph exists at \(x=1\). However, the simplified function \(x+2\) does not have this limitation and is continuous everywhere.

Step 4: Assess Equivalence of Functions

While \(x+2\) represents the simplified form of the original function, the two are not entirely equivalent. The original expression has a restriction at \(x=1\) where it is undefined, while \(x+2\) does not have this restriction. Therefore, these two functions are identical everywhere except at \(x=1\).

Key Concepts

Limits and Continuity

In mathematics, the concept of limits and continuity is fundamental when discussing the behavior of functions. A function is said to be continuous at a point if it is defined at that point and if its limit exists as it approaches that point. This means there's no sudden jump or gap in the function's value.

In our exercise, the original function \( \frac{(x-1)(x+2)}{x-1} \) is not defined at \( x = 1 \) because the denominator becomes zero, causing a discontinuity. Therefore, we have a 'hole' in the graph at this point. The function \( x+2 \), however, is continuous everywhere because it does not have any denominators or other factors that could become undefined.

It's important to check both the definition of a function at specific points and its behavior as it approaches those points when determining continuity. This helps us understand the overall nature of a function and any potential discontinuities.

Simplification of Rational Expressions

The simplification of rational expressions involves reducing a given fraction to its simplest form. This can be done by canceling out factors that appear in both the numerator and the denominator. Simplification makes expressions easier to work with but must be handled carefully.

In this exercise, the given function is \( \frac{(x-1)(x+2)}{x-1} \). Here, \( x-1 \) is a common factor that can be cancelled, simplifying the expression to \( x + 2 \). However, this simplification is only valid for values of \( x \) where the factor \( x-1 \) is not zero, ensuring that the original expression does not lead to division by zero.

When simplifying, it is crucial to consider the domain of the original expression, acknowledging values that cause the denominator to become zero. This ensures any changes to the expression still respect its original constraints.

Equivalence of Expressions

Two expressions are equivalent if they have the same value for all permissible values in their domain. When we simplify \( \frac{(x-1)(x+2)}{x-1} \) to \( x + 2 \), we achieve a form that is easier to manage, but are they truly equivalent?

Mathematically, the simplified expression \( x + 2 \) omits the restriction that the original expression had at \( x = 1 \). Since the original expression is undefined at this point due to a zero in the denominator, these two are not equivalent in terms of their domains.

We can say both expressions agree whenever \( x eq 1 \), but they differ exactly at \( x = 1 \). As the simplified expression \( x + 2 \) doesn’t acknowledge this point of discontinuity, they cannot be considered fully equivalent. This highlights the importance of considering the full domain and any restrictions that might apply when determining equivalence.