Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-15, state whether the indicated function is continu ous at 3. If it is not continuous, tell why. $$ r(t)= \begin{cases}\frac{t^{3}-27}{t-3} & \text { if } t \neq 3 \\ 23 & \text { if } t=3\end{cases} $$

Step-by-Step Solution

Answer

The function is not continuous at \( t = 3 \) because \( \lim_{t \to 3} r(t) = 27 \neq r(3) = 23 \).

Step 1: Evaluate the Function for Continuity Definition

A function is continuous at point \( c \) if three conditions are met: \( f(c) \) is defined, \( \lim_{x \to c} f(x) \) exists, and \( \lim_{x \to c} f(x) = f(c) \). We need to verify these conditions for \( t = 3 \).

Step 2: Check if \( r(3) \) is Defined

From the given piecewise function, \( r(t) \) is defined as 23 when \( t = 3 \). Therefore, \( r(3) = 23 \), and the first condition is satisfied.

Step 3: Calculate the Limit as \( t \to 3 \)

For the limit as \( t \to 3 \), we must consider the piece of the function where \( t eq 3 \):\[ \lim_{t \to 3} \frac{t^3 - 27}{t - 3} \]Simplify the expression by factoring the numerator:\[ \frac{t^3 - 27}{t - 3} = \frac{(t - 3)(t^2 + 3t + 9)}{t - 3} \]Cancel out \( t - 3 \) from the numerator and denominator:\[ \lim_{t \to 3} t^2 + 3t + 9 = 3^2 + 3(3) + 9 = 27 \].

Step 4: Compare the Limit to \( r(3) \)

The limit as \( t \) approaches 3 is 27, while \( r(3) = 23 \). Since these two values are different, \( \lim_{t \to 3} r(t) eq r(3) \), and thus the function is not continuous at \( t = 3 \).

Key Concepts

Limit of a Function

Understanding the limit of a function is crucial in calculus, as it describes the behavior of a function as it approaches a certain point. When analyzing a function to determine its continuity at a point, the limit helps us see if the left-hand and right-hand limits agree with each other and with the function's value at that point.

To check if a function is continuous at a point, you need to ensure three things:

• The function value exists at that point.
• The limit of the function exists as it approaches that point, meaning both the left-hand and right-hand limits must agree.
• The function's value equals the calculated limit at that point.

The given exercise provides us with a function, where we need to evaluate the limit as it approaches 3. In our specific case, by using algebraic techniques like factoring, we can simplify the expression and find the limit. For example, the expression \( \frac{t^3 - 27}{t - 3} \) gets simplified through factorization, allowing us to determine the limit as \( t \) approaches 3, which is 27. However, as the function value at \( t = 3 \) is 23, we find that even though the limit exists and is well-defined mathematically, it does not match the function value, indicating a discontinuity.

Piecewise Functions

Piecewise functions are a common concept in calculus, where a function is defined using different expressions depending on the domain we're looking at. A single function definition can't always capture all types of behavior, and piecewise definitions step in to represent function behavior over certain intervals.

In the provided problem, the function \( r(t) \) is a textbook example. It is defined by two separate pieces:

• \( \frac{t^3 - 27}{t - 3} \) for \( t eq 3 \)
• 23 for \( t = 3 \)

This duality in definition allows the function to reflect different behaviors at different points, but it also makes finding continuity at specific points more complex.

When assessing continuity for a piecewise function, you need to evaluate each piece around the point of interest and ensure they agree at the exact point. If they don't, as in this example at \( t = 3 \), where the limit and the function value do not match even though the individual expressions are well-defined, the piecewise function is not continuous at that point.

Factorization in Algebra

Factorization is an algebraic technique used to simplify expressions, especially when dealing with fractions or polynomial equations. In this exercise, factorization is employed to simplify the given function \( \frac{t^3 - 27}{t - 3} \).

This expression might look intimidating at first, but it can be simplified by recognizing the numerator as a difference of cubes. The formula for a difference of cubes is \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \).

Applying this formula, we can rewrite \( t^3 - 27 \) as \((t - 3)(t^2 + 3t + 9)\). This allows for the cancellation of \( t - 3 \) from the numerator and denominator, simplifying the limit expression to \( t^2 + 3t + 9 \) as \( t \to 3 \).

Using factorization not only makes the process of finding limits easier but also reduces the complexity of the expression, enabling you to handle and interpret it more effectively. This illustrates how mastering basic algebraic methods can significantly aid in solving calculus problems.