Problem 29 - Calculus: Early Transcendentals

Question

Continuity Determine the interval(s) on which the following functions are continuous. $$f(x)=\frac{1}{x^{2}-4}$$

Step-by-Step Solution

Answer

Answer: The function is continuous on the intervals $$(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$$

Step 1: Identify the discontinuity points

Find the values of x for which the denominator is zero: $$x^2 - 4 = 0$$ Solve this quadratic equation for x.

Step 2: Solve the quadratic equation

To solve the quadratic equation, factor it: $$(x + 2)(x - 2) = 0$$ Now, set each factor equal to zero, and solve for x: $$x + 2 = 0 \Rightarrow x = -2$$ $$x - 2 = 0 \Rightarrow x = 2$$ The points of discontinuity are x = -2 and x = 2.

Step 3: Determine the intervals of continuity

Since the function is continuous everywhere except at the points x = -2 and x = 2, the intervals of continuity are the intervals that do not include these points. We have three intervals of continuity: 1. \((-\infty, -2)\) 2. \((-2, 2)\) 3. \((2, \infty)\)

Step 4: Write the final answer

The function $$f(x)=\frac{1}{x^{2}-4}$$ is continuous on the intervals: $$(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$$

Key Concepts

Intervals of Continuity

In mathematics, a function is said to be continuous on an interval if there are no breaks, jumps, or holes in its graph throughout that interval. For the given function \(f(x) = \frac{1}{x^2 - 4}\), we are interested in finding the intervals where this function remains smooth and uninterrupted.

To do this, we look for the values of \(x\) where the function could potentially break or become undefined. This happens primarily where the denominator of the function becomes zero, as dividing by zero is undefined. Once we find these points, we identify the continuous intervals as being between these discontinuities, but not including them.

The exercise tells us that the points \(x = -2\) and \(x = 2\) are such discontinuities. Therefore, the function will be continuous in the intervals that do not contain these points:

• The interval \((-fty, -2)\)
• The interval \((-2, 2)\)
• The interval \((2, fty)\)

It is in these stretches that the function behaves in a predictable and smooth manner, free from sudden changes or gaps in the graph.

Discontinuity Points

Discontinuity points are those specific values of \(x\) where a function like \(f(x) = \frac{1}{x^2 - 4}\) becomes undefined or breaks in some way. For rational functions, discontinuities frequently occur where the denominator equals zero. This is important because division by zero is undefined in mathematics, leading to a gap or asymptote at these points.

In the equation \(x^2 - 4 = 0\), solving for \(x\) gives the points of discontinuity. As outlined in the solution, factoring leads to \((x+2)(x-2) = 0\) giving two solutions:

• \(x = -2\)
• \(x = 2\)

These solutions correspond to the locations on the \(x\)-axis where the function suddenly becomes non-existent or changes drastically. Graphically, you might see these as vertical gaps or lines not touching the \(x\)-axis at those exact points. Understanding where these points of discontinuity occur helps in graphing the function accurately and understanding its behavior on different intervals.

Quadratic Equation

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The one we are dealing with here, \(x^2 - 4 = 0\), is a classic example, albeit a relatively simple one, where \(b = 0\) and \(c = -4\).

Solving quadratic equations like this one can be done by various methods, including factoring, completing the square, or using the quadratic formula. In this context, factoring is the most straightforward approach:

• Rewrite \(x^2 - 4\) as \((x+2)(x-2)\).
• Set each factor equal to zero to find the solutions: \(x = -2\) and \(x = 2\).

This process shows the advantage of factoring, clearly breaking down the equation into parts that set the path to finding its roots or solutions. Not only do these solutions help in addressing the quadratic itself, but they also reveal discontinuity points in functions that contain the equation in their denominator.