Question

Determine whether the function is continuous on the entire real line. Explain your reasoning. \(f(x)=\frac{1}{x^{2}-4}\)

Short answer

The function \(f(x)=\frac{1}{x^{2}-4}\) is continuous on the real line except at \(x=-2\) and \(x=2\). The intervals of continuity are \((-∞, -2)\), \((-2, 2)\), and \((2, ∞)\).

Step-by-step solution

Identify the discontinuity points

First, locate where the function \(f(x)=\frac{1}{x^{2}-4}\) becomes undefined. It becomes undefined when the denominator equals zero. So, solve the equation \(x^{2}-4=0\) for \(x\).

Solve the equation

Solving \(x^{2}-4=0\) gives \(x^{2}=4\), which implies \(x=\pm2\). The function \(f(x)\) is therefore undefined at \(x=-2\) and \(x=2\).

Establish the intervals of continuity

Since the function is undefined only at \(x=-2\) and \(x=2\), it is continuous everywhere else on the real line. Therefore, the intervals of continuity of \(f(x)\) are \((-∞, -2)\), \((-2, 2)\) and \((2, ∞)\).

Key concepts

Discontinuity Points

Discontinuity points in a function occur where the function is not continuous. These are specific values of the input where the function suddenly jumps, breaks, or stops being defined. In the provided example, the goal is to determine these points for the function \( f(x)=\frac{1}{x^{2}-4} \).
To find such points, you must identify where the denominator of the fraction equals zero, as a division by zero is undefined. Thus, solving the equation \( x^{2}-4=0 \) allows us to pinpoint these critical values. The solution to \( x^{2}=4 \) reveals two points: \( x=2 \) and \( x=-2 \).
At these values, the denominator becomes zero, rendering the function discontinuous at \( x=2 \) and \( x=-2 \). As a result, these are the discontinuity points for the given function.

Undefined Function

When talking about an undefined function in a calculus context, it refers to places where a function fails to produce a value. This generally happens when an operation within the function can't be mathematically satisfied.
In our example, the function \( f(x)=\frac{1}{x^{2}-4} \) becomes undefined when \( x^{2}-4 \) equals zero due to division by zero. This occurs specifically at \( x=2 \) and \( x=-2 \).
Whenever the denominator in a fraction equals zero, the function is undefined at those x-values because division by zero is a mathematical operation that has no solution. Hence, the function does not exist at these points, which are crucial for understanding the behavior and limitations of the function.

Intervals of Continuity

Intervals of continuity describe the range of x-values where a function is continuous, avoiding any undefined points or breaks. For continuous intervals, you don't encounter any holes, jumps, or asymptotic behavior.
In the case of the function \( f(x)=\frac{1}{x^{2}-4} \), continuity is interrupted only at \( x=-2 \) and \( x=2 \), where the function is undefined. Consequently, the function remains continuous on the following intervals:

• \((-∞, -2)\)
• \((-2, 2)\)
• \((2, ∞)\)

These intervals collectively express where you can expect the function to behave predictably and without interruption, except at the aforementioned points of discontinuity.