Question

Determine the interval(s) on which the following functions are continuous. Be sure to consider right-and left-continuity at the endpoints. $$f(t)=\left(t^{2}-1\right)^{3 / 2}$$

Short answer

Answer: The function is continuous on the interval $$t \leq -1 \: \text{or} \: t \geq 1.$$

Step-by-step solution

Identify the domain of the function

The given function is $$f(t)=\left(t^{2}-1\right)^{3 / 2}.$$ We need to make sure that the radicand \((t^2 - 1)\) is non-negative, since taking a square root of a negative number returns a complex number. So, we have the condition $$t^{2} - 1 \geq 0.$$ Solving for \(t\), we find the domain of the function to be $$t \leq -1 \: \text{or} \: t \geq 1.$$

Check right and left continuity at the endpoints

We have two endpoints for the domain of \(f(t)\): \(t = -1\) and \(t = 1\). Let's analyze the left and right continuity at these points. At \(t = -1\), we only need to check for the right continuity, since the function is not defined for values less than -1. Right continuity at \(t = -1\) implies that $$\lim_{t \rightarrow -1^+} f(t) = f(-1).$$ Now, let's evaluate the limit and the function at \(t = -1\): $$\lim_{t \rightarrow -1^{+}}\left( t^2 - 1 \right)^{3/2} = \left((-1)^2 - 1 \right)^{3/2} = 0^{3/2} = 0.$$ Since both the limit and the function value are equal, right continuity holds at \(t = -1\). At \(t = 1\), we only need to check for left continuity, since the function is not defined for values greater than 1. Left continuity at \(t = 1\) implies that $$\lim_{t \rightarrow 1^-} f(t) = f(1).$$ Now, let's evaluate the limit and the function at \(t = 1\): $$\lim_{t \rightarrow 1^{-}}\left( t^2 - 1 \right)^{3/2} = \left((1)^2 - 1 \right)^{3/2} = 0^{3/2} = 0.$$ Since both the limit and the function value are equal, left continuity holds at \(t = 1\).

State the interval(s) on which the function is continuous

Since the left and right continuity hold at the endpoints, and the function is continuous on its domain, we can say that the function \(f(t) = \left(t^{2}-1\right)^{3 / 2}\) is continuous on its entire domain, which is $$t \leq -1 \: \text{or} \: t \geq 1.$$

Key concepts

Intervals of Continuity

The intervals of continuity for a function tell us where the function is continuous, meaning there are no breaks, jumps, or holes in the graph. For the given function, \(f(t) = (t^2 - 1)^{3/2}\), we want it to be continuous within the intervals where it's defined.
This means checking:

• Whether the function includes complete, unbroken intervals across its domain.
• If there are any points or intervals in the range of \(t\) where the function "jumps" or is undefined.

For \(f(t)\), the key is making sure the part under the square root, \(t^2 - 1\), is non-negative, because otherwise the expression would not be a real number. The solution for \(t^2 - 1 \geq 0\) leads us to intervals \(t \leq -1\) or \(t \geq 1\).
Therefore, the function is continuous on these intervals since it satisfies all conditions of unbroken or seamless connection within these ranges.

Domain of a Function

Understanding the domain of a function is foundational in determining its intervals of continuity. The domain is simply all the possible input values \(t\) that make the function work out to a real number.
For \(f(t) = (t^2 - 1)^{3/2}\), the critical instruction provided was ensuring \(t^2 - 1\) stays non-negative, resulting in \(t \leq -1\) or \(t \geq 1\). These limits describe which numbers you can safely insert into \(t\) without causing math headaches like square roots of negatives, which aren't real numbers.
In other words, the domain consists of two disjoint intervals:

• All numbers less than or equal to -1.
• All numbers greater than or equal to 1.

Ensuring these conditions keeps the function within the realm of real numbers and helps map its continuous behavior. Domains bridge the gap to analyzing both the overall shape and behavior of a function.

Left and Right Continuity

Left and right continuity are concepts that help examine function behavior at specific points, particularly at boundaries of domains. A function is left continuous at a point if a small move leftwards doesn’t disrupt its continuity. Right continuity follows similarly for rightward moves.
For \(f(t)\), we focus on points \(t = -1\) and \(t = 1\):

• At \(t = -1\), we check right continuity because the function is not defined for values less than -1.
• Right continuity at \(t = -1\) involves ensuring the limit from the right, \(\lim_{t \to -1^+} f(t)\), matches the function value \(f(-1)\), which both equal zero in this case.

Similarly,

• At \(t = 1\), we check left continuity, as the function is undefined for values greater than 1.
• Left continuity at \(t = 1\) requires the limit from the left, \(\lim_{t \to 1^-} f(t)\), matches \(f(1)\), which again results in zero for both.

By ensuring both left and right continuity at endpoints, we verify the function is seamlessly connected in its defined domain, making these points as smooth as the rest of the function's journey across its intervals.