Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

State whether the indicated function is continuous at \(3 .\) If it is not continuous, tell why. $$ f(t)=\left\\{\begin{array}{cl} t^{2}-9 & \text { if } t \leq 3 \\ (3-t)^{2} & \text { if } t>3 \end{array}\right. $$

Step-by-Step Solution

Answer

The function is continuous at \( t = 3 \).

Step 1: Check if Function is Defined at the Point

To check continuity at a point (here, at \( t = 3 \)), the first step is to ensure the function is defined at that point. The given function is piecewise: for \( t \leq 3 \), the function is \( f(t) = t^2 - 9 \). Therefore, \( f(3) = 3^2 - 9 = 0 \). The function is defined at \( t = 3 \).

Step 2: Determine the Left-Hand Limit as t Approaches 3

For the left-hand limit, we examine the part of the function where \( t \leq 3 \). Here, \( f(t) = t^2 - 9 \). Compute the limit as \( t \) approaches 3 from the left:\[\lim_{{t \to 3^-}} (t^2 - 9) = 3^2 - 9 = 0.\]

Step 3: Determine the Right-Hand Limit as t Approaches 3

For the right-hand limit, we need the part of the function where \( t > 3 \). Here, \( f(t) = (3 - t)^2 \). Compute the limit as \( t \) approaches 3 from the right:\[\lim_{{t \to 3^+}} (3-t)^2 = (3-3)^2 = 0.\]

Step 4: Compare Limits and Function Value

Continuity at a point requires the left-hand limit, right-hand limit, and function value at that point to be equal. Here, \( \lim_{{t \to 3^-}} f(t) = 0 \), \( \lim_{{t \to 3^+}} f(t) = 0 \), and \( f(3) = 0 \). Since all are equal, \( f(t) \) is continuous at \( t = 3 \).

Key Concepts

Piecewise Function

A piecewise function is a function that is defined by different formulas or expressions, depending on the value of the variable. Rather than having a single rule for all inputs, it has multiple rules, each applicable to a certain interval of inputs. This makes them very versatile and useful for modeling situations where conditions change. Here, our function is divided based on the value of variable \( t \).

• If \( t \leq 3 \), the function behaves as \( f(t) = t^2 - 9 \).
• If \( t > 3 \), the function behaves as \( f(t) = (3-t)^2 \).

Understanding how to handle piecewise functions in calculations and graphing is crucial for solving continuity problems.

Left-hand Limit

The left-hand limit of a function at a point refers to the value that the function approaches as the variable gets arbitrarily close to the point from the left (i.e., from values less than the point). It's written as \( \lim_{{t \to 3^-}} f(t) \). Calculating this helps us predict the behavior of the function just before reaching the specified point from one direction.
For our piecewise function, as \( t \) approaches 3 from the left, we use the segment \( f(t) = t^2 - 9 \). Evaluating the limit:

• \( \lim_{{t \to 3^-}} (t^2 - 9) = 3^2 - 9 = 0 \).

The limit shows that as we get very close to 3 from the left, the function gets very close to 0.

Right-hand Limit

Similarly, the right-hand limit indicates the value a function approaches as the variable comes from the right-hand side, i.e., from values greater than the point. It is written as \( \lim_{{t \to 3^+}} f(t) \). Knowing this helps determine if the behavior of the function is consistent on either side of the point.
For our function, as \( t \) approaches 3 from the right, we use the expression \( f(t) = (3-t)^2 \). Calculating this limit:

• \( \lim_{{t \to 3^+}} (3-t)^2 = (3-3)^2 = 0 \).

Thus, from the right side, the function also approaches 0, showing similar behavior to the left side.

Function Evaluation

Function evaluation at a specific point means determining the output of the function directly at that point. This step is crucial for checking the continuity of a function at that point. For continuity, not only should the left-hand and right-hand limits be equal, they should also equal the function's value at the point.
In our piecewise function, evaluating \( f(t) \) at \( t = 3 \) gives us:

• For \( t \leq 3 \), \( f(3) = 3^2 - 9 = 0 \).

Function evaluation, combined with checking both the left and right-hand limits, showed us that at \( t = 3 \), the function value is 0. Since the left and right limits and the value at the point are all equal, the function is continuous at \( t = 3 \).