Problem 12 - 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. \(f(x)=\frac{1}{x^{2}-4}\)

Step-by-Step Solution

Answer

The function \(f(x)=\frac{1}{x^{2}-4}\) is continuous on the intervals \(x<-2\), \(-22\). It is not continuous at \(x=-2\) and \(x=2\) which are the points of discontinuity as the denominator becomes zero leading to division by zero, which violates the conditions for continuity.

Step 1: Find the Points of Discontinuity

Set the denominator of the function equal to zero and solve for \(x\). \nTherefore, \(x^{2}-4=0\). Applying the difference of squares formula, we get \((x+2)(x-2)=0\). The solutions are \(x=-2, 2\), which are the points of discontinuity.

Step 2: Determine the Intervals of Continuity

Now, we establish the intervals of continuity which are basically the entire real number line except the points of discontinuity. \nThe intervals of continuity are \(x<-2\), \(-22\).

Step 3: Explaining the Continuity

The function \(f(x)=\frac{1}{x^{2}-4}\) is a rational function, and rational functions are continuous everywhere except for where their denominator is 0 (since division by 0 is undefined). The points \(x=-2\) and \(x=2\) are the only cases where the denominator becomes 0, causing discontinuities. Therefore, except for these points, the function is continuous for all other real numbers.

Step 4: Identifying the conditions of continuity not satisfied

At \(x=-2\) and \(x=2\), the function isn't continuous because it doesn't meet the following condition for continuity: A function \(f(x)\) is continuous at \(x=a\) if and only if \(\lim_{x\to a^-}f(x) = f(a) = \lim_{x\to a^+}f(x)\). At \(x=-2\) and \(x=2\), \(f(x)\) is not defined so this condition is not satisfied.

Key Concepts

Rational Functions

Rational functions are forms of mathematical expressions comprising two polynomials, one being the numerator and the other as the denominator. They take the form \( f(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
Rational functions are broadly used due to their versatility in expressing complex relationships and are particularly useful in calculus and algebra.

• A prime feature of rational functions is their continuity, which generally holds true across their domains.
• The only places where rational functions are not continuous are where the denominator, \( Q(x) \), equals zero.

To further illustrate, consider the function \( f(x) = \frac{1}{x^2 - 4} \). Here, the numerator is 1 (a constant polynomial), and the denominator is \( x^2 - 4 \). Where the denominator becomes zero, points of discontinuity arise, indicating places where the function is not defined.

Points of Discontinuity

Points of discontinuity refer to values of \( x \) where a function is not continuous, effectively resulting in a break in the graph of the function. This occurs in rational functions where the denominator is zero.
Let's unravel the points of discontinuity in \( f(x) = \frac{1}{x^2 - 4} \):

• Set the expression in the denominator equal to zero: \( x^2 - 4 = 0 \).
• Solve for \( x \) to find critical points: \( (x + 2)(x - 2) = 0 \).
• The solutions, \( x = -2 \) and \( x = 2 \), identify where discontinuities occur.

As these values cause the denominator to vanish, the function \( f(x) \) is undefined at these specific points. However, the function remains defined and continuous on intervals apart from these discontinuities.

Conditions of Continuity

The conditions for a function \( f(x) \) to be continuous at a particular point \( x = a \) can be summarized as follows:

• The function \( f(a) \) must be defined.
• The limit \( \lim_{x \to a} f(x) \) must exist.
• The limit value must equal the function value: \( \lim_{x \to a} f(x) = f(a) \).

Let's apply these conditions to the function \( f(x) = \frac{1}{x^2 - 4} \). At \( x = -2 \) and \( x = 2 \), the function is not continuous because:- The denominator becomes zero, so the function \( f(x) \) is not defined.- Consequently, since \( f(x) \) is undefined at these points, the other conditions of continuity aren't met either.
By identifying these points, one can precisely determine the intervals where \( f(x) \) is continuous, which happens to be every real number except \( x = -2 \) and \( x = 2 \).