Problem 20 - Calculus: Early Transcendentals

Question

Explain why the function is discontinuous at the given number \( a \). Sketch the graph of the function. \( f(x) = \left\\{ \begin{array}{ll} \dfrac{x^2 - x}{x^2 - 1} & \mbox{if \) x \neq 1 \(} \hspace{40mm} a = 1\\\ 1 & \mbox{if \) x = 1 \(} \end{array} \right.\)

Step-by-Step Solution

Answer

The function is discontinuous at \( x = 1 \) because \( \lim_{x \to 1} f(x) = \frac{1}{2} \) and \( f(1) = 1 \).

Step 1: Analyze the Given Function

The function \( f(x) \) has two expressions. When \( x eq 1 \), \( f(x) = \frac{x^2 - x}{x^2 - 1} \). When \( x = 1 \), \( f(x) = 1 \). We'll check if these two expressions yield the same value as \( x \) approaches 1.

Step 2: Simplify the Expression for \( x \neq 1 \)

For \( x eq 1 \), we simplify \( \frac{x^2 - x}{x^2 - 1} \). Factor the numerator as \( x(x-1) \) and the denominator as \( (x-1)(x+1) \). The function simplifies to \( \frac{x}{x+1} \) when \( x eq 1 \).

Step 3: Evaluate the Limit as \( x \to 1 \)

Calculate \( \lim_{x \to 1} \frac{x}{x+1} \). Substituting \( x = 1 \), we get \( \frac{1}{1+1} = \frac{1}{2} \). This is the value that \( f(x) \) approaches as \( x \to 1 \), which is not equal to \( f(1) = 1 \).

Step 4: Determine Discontinuity at \( a = 1 \)

A function is discontinuous at a point if the limit of the function as \( x \) approaches the point is not equal to the function's value at that point. Since \( \lim_{x \to 1} f(x) = \frac{1}{2} \) and \( f(1) = 1 \), \( f(x) \) is discontinuous at \( x = 1 \).

Step 5: Sketch the Graph of the Function

Draw a graph representing \( f(x) = \frac{x}{x+1} \) for \( x eq 1 \), which is a hyperbola. Mark a hole at \( x = 1 \) because the function is not defined there. Plot the point \( (1, 1) \) to represent \( f(1) = 1 \). The graph on both sides of \( x = 1 \) approaches the line \( y = \frac{1}{2} \), illustrating the discontinuity.

Key Concepts

Limit Analysis

In mathematics, analyzing the limit of functions helps us understand their behavior as the input approaches a particular value. For the function given in the exercise, we need to analyze the limit as \( x \) approaches 1. This function has two different expressions depending on whether \( x \) is equal to 1 or not.
For \( x eq 1 \), the expression is \( \frac{x^2 - x}{x^2 - 1} \). By simplifying this, we derive \( \frac{x}{x+1} \). As \( x \to 1 \), employing direct substitution in this simplified expression, we observe:

• When we set \( x \to 1 \), the expression \( \frac{x}{x+1} \) yields a limit of \( \frac{1}{2} \).

This value gives us the expected behavior or trend of the function around \( x = 1 \), without actually touching \( x = 1 \). However, we've defined \( f(1) = 1 \), which creates a discrepancy. Therefore, the function does not match its trend at \( x = 1 \). This is a classic indication of discontinuity.

Function Graph Sketch

Sketching a function graph provides a visual representation of its behavior and helps identify points of interest such as discontinuities. For the given problem, we handle \( f(x) = \frac{x}{x+1} \) for \( x eq 1 \), which portrays a hyperbolic curve.
To sketch the graph:

• Draw the curve of \( \frac{x}{x+1} \).
• Place a hole at \( x = 1 \) (since the simplified function is not defined there).
• Plot the point \( (1, 1) \) because \( f(x) \), at \( x = 1 \), outputs 1 due to the predefined value.

Understanding the layout of the graph with these elements clearly signifies the discontinuity seen just by the visualization of the hole versus the point plotted at \( (1,1) \).

Factorization

Factorization is the process of breaking down expressions into multiplicable components, making complex functions simpler to analyze. The discontinuous function \( f(x) = \frac{x^2 - x}{x^2 - 1} \) requires factorization for \( x eq 1 \). By factoring:

• The numerator, \( x^2 - x \), can be rewritten as \( x(x-1) \).
• The denominator, \( x^2 - 1 \), can be rewritten as \( (x-1)(x+1) \).

This results in the expression \( \frac{x(x-1)}{(x-1)(x+1)} \). We can simplify by canceling the common term \((x-1)\) in both the numerator and denominator as long as \( x eq 1 \), leaving us with \( \frac{x}{x+1} \).
Through factorization, the true nature of the function near \( x = 1 \) is revealed, which contributes greatly to determining its limit and visual behavior.

Function Value Comparison

By comparing function values at specific points with their trend approaching that point, we identify discontinuities. For the exercise at hand, focus on the behavior of \( f(x) \, as \, x \to 1 \):

• For any \( x eq 1 \), the function simplifies to \( \frac{x}{x+1} \), approaching a limiting value of \( \frac{1}{2} \).
• However, we manually set \( f(1) = 1 \).

Discontinuity occurs because the trend seen through the limit \( \frac{1}{2} \) does not match the actual function value \( 1 \) at \( x = 1 \). This comparison of values unveils the gap between the continuous approach and the jump at the function's predefined point.
In conclusion, understanding these disparities showcases the essence of a discontinuous function, making this comparison an effective tool for learners.