Problem 26 - Calculus Early Transcendentals

Question

Use Theorem 2.10 to determine the intervals on which the following functions are continuous. $$f(t)=\frac{t+2}{t^{2}-4}$$

Step-by-Step Solution

Answer

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

Step 1: Identify points of discontinuity

To find the points of discontinuity, we must identify the values of \(t\) where the denominator is equal to zero. In this case, the denominator is \(t^{2}-4\). So, we need to solve the equation: $$t^{2}-4=0$$ Factor the equation as a difference of squares: $$(t-2)(t+2)=0$$ We can now find the two points of discontinuity: $$t=2, \quad t=-2$$

Step 2: Construct the intervals of continuity

Now that we have found the points of discontinuity, we can construct the intervals of continuity. In this case, the function is continuous for all values of \(t\) that do not cause the denominator to become zero, so the intervals of continuity are: $$(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$$ The function \(f(t)=\frac{t+2}{t^{2}-4}\) is continuous on the intervals \((-\infty, -2)\), \((-2, 2)\), and \((2, \infty)\).

Key Concepts

Intervals of Continuity

When discussing the intervals of continuity for a function, we refer to the ranges within which the function can be graphed without lifting a pencil. This is a way of visualizing that there are no breaks or holes in the graph for those specific ranges.

For the given function \(f(t)=\frac{t+2}{t^{2}-4}\), we've identified points where the function cannot be continuous - where the denominator equals zero. What's left are the intervals \( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \). These intervals tell us that if we pick any number within these ranges, we'll get a real, finite number output from our function. There's no need to worry about division by zero in these ranges. While some functions are continuous on all real numbers, it's the interruption at these critical points that defines our intervals of continuity.

Points of Discontinuity

When we examine a function like \(f(t)=\frac{t+2}{t^{2}-4}\), not every value of \(t\) will yield a well-behaved function. The points of discontinuity are the specific values that cause the function to 'misbehave', typically leading to undefined or infinite results. Here, if \(t\) is 2 or -2, our denominator reduces to zero, and we cannot divide by zero in standard mathematics. This creates what we call singularities or points of discontinuity at \(t=2\) and \(t=-2\).

To identify such points, we look for where the function's denominator is zero, as the absence of a value in the denominator signifies that the function does not exist at those points. These are crucial to understanding the overall behavior of the function as they are often associated with holes, jumps, or vertical asymptotes on the graph.

Difference of Squares

A difference of squares refers to an algebraic expression of the form \( a^2 - b^2 \), which can be factored into \( (a + b)(a - b) \). This is a useful technique for solving equations that have the structure of a squared term subtracted from another squared term.

In our example with the function \(f(t)=\frac{t+2}{t^{2}-4}\), the denominator \( t^2 - 4 \) is a difference of squares because \(4\) is a perfect square \( (2^2)\) and \(t^2\) is the square of \(t\). By factoring the denominator as \( (t-2)(t+2)\), we can more easily identify the points of discontinuity at \(t=2\) and \(t=-2\). Understanding how to factor and utilize the property of the difference of squares is crucial for solving rational functions, especially when determining their domains and discontinuities.