Problem 18 - College Algebra and Calculus: An Applied Approach

Question

Describe the interval(s) on which the function is continuous. Explain why the function is continuous on the interval(s). If the function has a discontinuity, identify the conditions of continuity that are not satisfied. \(f(x)=\frac{x-3}{x^{2}-9}\)

Step-by-Step Solution

Answer

The function \(f(x)=\frac{x-3}{x^{2}-9}\) is continuous on \(x \in (-\infty, -3) \cup (-3, 3) \cup (3, \infty)\). It has points of discontinuity at \(x = -3\) and \(x = 3\), where the conditions of continuity are not met due to non-existence of limit at these points.

Step 1: Look at the function and find its domain

The function \(f(x)=\frac{x-3}{x^{2}-9}\) can be simplified as \(f(x)=\frac{x-3}{(x-3)(x+3)}\). Now, you can simplify it further to \(f(x)=\frac{1}{x+3}\), except when \(x = 3\), because \(x = 3\) is not in the domain of original function. The function will be defined at every point except for \(x = -3\) and \(x = 3\). So, we already found two points of discontinuity.

Step 2: Test the limit at the point of discontinuity

One of the conditions of continuity is that the limit of the function exists at the point. So if the limit exists at \(x = -3\) and \(x = 3\), we have a case of removable discontinuity. However, if the limit does not exist, we have a non-removable or essential discontinuity. Here in this case, limit does not exist at \(x = -3\) and \(x = 3\). So, these points are a case of non-removable or essential discontinuity.

Step 3: Conclude and list intervals of continuity

Knowing the points of discontinuity, it is now possible to determine the intervals on which the function is continuous. As it is discontinuous at \(x = -3\) and \(x = 3\) and continuous everywhere else, it is continuous on \(x \in (-\infty, -3) \cup (-3, 3) \cup (3, \infty)\)

Key Concepts

Discontinuity Types

In calculus, understanding the types of discontinuities in functions is crucial. Discontinuities are points where a function is not continuous.
There are several types of discontinuities:

• Removable Discontinuity: Occurs when a hole exists in the graph, and the function can be fixed by assigning the value of the limit at that point.
• Jump Discontinuity: Happens when there is a sudden "jump" in the values of the function at a certain point, where the left-hand and right-hand limits exist but are not equal.
• Infinite Discontinuity: Takes place when the function approaches infinity as it nears the discontinuous point.

For the function given, at points \(x = -3\) and \(x = 3\), we have essential or non-removable discontinuities. The limits do not exist, which means these points cannot be "fixed" by merely redefining function values at those points.

Domain of a Function

The domain of a function is the set of all possible input values (\(x\)-values) that a function can accept. It defines where the function is valid and can be evaluated.
For the function \(f(x) = \frac{x-3}{x^2-9}\), we first factor the denominator to observe restrictions in the domain.
After simplifying, \(f(x) = \frac{1}{x+3}\), except \(x=3\) is excluded due to the original formulation. The values \(x = -3\) and \(x = 3\) make the denominator zero, which means the function is undefined at these points.

• To find the domain, exclude these points from the real number line.
• The valid \(x\)-values for this function are \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\).

Understanding the domain helps to identify which intervals need further examination for continuity.

Intervals of Continuity

Intervals of continuity refer to the parts of the domain where a function is continuous without any breaks or holes. For a function to be continuous on an interval:

• The function must be defined at every point in the interval.
• The limits must exist and equal the function's value at every point within the interval.

In the problem presented, since the function \(f(x) = \frac{x-3}{x^2-9}\) is discontinuous only at \(x = -3\) and \(x = 3\), it is continuous everywhere else. It means:

• The intervals of continuity are \((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\).

Knowing these intervals allows us to confidently work with the function in those zones, understanding it behaves predictably without interruptions.