Question

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

Short answer

Question: Determine the interval(s) on which the function \(f(t)=\frac{6t}{t^2-4}\) is continuous. Answer: The function is continuous on the intervals \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\).

Step-by-step solution

Identify problematic points

For a function to be continuous, the denominator must not be equal to zero. So we need to find the values of \(t\) for which the denominator is equal to zero. Rewrite the function and set the denominator equal to zero: $$t^{2}-4=0$$

Solve for problematic points

Now, we need to solve the equation for the values of \(t\): $$(t-2)(t+2)=0$$ This gives us two solutions: $$t=2, \quad t=-2$$

Continuity evaluation

The function can only be discontinuous at the problematic points found in step 2. Now we will determine if the function is continuous or discontinuous at these points. For \(t=2\), we evaluate the limit from the left and right: $$\lim_{t\rightarrow 2^-} f(t) = \lim_{t\rightarrow 2^+} f(t)$$ Similarly, for \(t=-2\): $$\lim_{t\rightarrow -2^-} f(t) = \lim_{t\rightarrow -2^+} f(t)$$

Identify the interval(s) of continuity

Since our function is continuous, except for the problematic points \(t=2\) and \(t=-2\). The function is discontinuous at these points because the denominator becomes zero. So the interval of continuity will be: $$(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$$

Key concepts

Limits of Functions

The concept of a limit is fundamental when we're talking about continuity in functions. It helps us understand the behavior of a function as it approaches a specific point, but not necessarily when it reaches that point. In other words, when you're calculating the limit of a function at a certain value, you're basically looking at what value the function is heading towards—not where it actually lands.

For instance, in the provided exercise with the function \( f(t)=\frac{t+2}{t^2-4} \), we consider what happens as \( t \) approaches the values where the denominator equals zero. If the limits from both the left (indicated by \( t \) approaching a value from the negative side, or \( t^- \)) and right (indicated by \( t \) approaching a value from the positive side, or \( t^+ \)) are the same, we say the limit exists. Otherwise, the function has a discontinuity at that point. This understanding is crucial to determining the intervals on which a function is continuous.

Rational Functions

A rational function, such as \( f(t)=\frac{t+2}{t^2-4} \) from our exercise, is characterized by having both a numerator and a denominator that are polynomials. One of the main properties of these functions is that they can be discontinuous at certain points—specifically at the values that make the denominator equal to zero.

In the step-by-step solution, the first task is to identify such points by setting the denominator equal to zero, which gives us the problematic points where the function could potentially be discontinuous. It's important to note that while a rational function may not be defined at these points, it can approach finite limits as long as the numerators do not also become zero at the same points. This is essential for determining continuity, as we need to understand where the function does not have a well-defined value.

Interval of Continuity

An interval of continuity refers to a region on the number line over which a function is continuous. For a function to be continuous on an interval, it must be defined for every point within that interval, and have matching left-hand and right-hand limits at every point within the interval.

In our exercise, we found that the function \( f(t)=\frac{t+2}{t^2-4} \) is discontinuous at \( t=2 \) and \( t=-2 \) because at these points, the denominator becomes zero. By excluding these points, we can say the function is continuous on the intervals \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \). There is no single continuous interval since we have to break it up at the points where the function cannot be defined. Recognizing these intervals is important for a variety of calculus problems, including integration and finding limits at infinity.