Problem 40 - Calculus: Graphical, Numerical, Algebraic

Question

Group Activity In Exercises \(37-40\) , verify that the function is continuous and state its domain. Indicate which theorems you are using, and which functions you are assuming to be continuous. $$y=\left\\{\begin{array}{ll}{\frac{x^{2}-1}{x-1},} & {x \neq 1} \\ {2,} & {x=1}\end{array}\right.$$

Step-by-Step Solution

Answer

Yes, the function is continuous for all real numbers which also makes up its domain.

Step 1: Identify the Domain

By examining the function, it can be seen that it has two parts. The first part of the function \(\frac{x^{2}-1}{x-1}\) is defined for all \(x \neq 1\). The second part of the function is defined only for \(x = 1\). Thus, the domain of the function is all real numbers.

Step 2: Check Continuity for \(x \neq 1\)

The first part of the function is a quotient of two functions that are known to be continuous since they are polynomials. According to the continuous function theorem, the quotient of two continuous functions \(f(x) / g(x)\) is continuous at all points where \(g(x) \neq 0\). Here, \(g(x)\) = \(x-1\), which is not equal to zero for all \(x \neq 1\), so it is continuous everywhere except at \(x = 1\).

Step 3: Check Continuity for \(x = 1\)

Looking at the second part of the function which is \(y=2\) for \(x = 1\), it is a constant function and constant functions are always continuous. Thus, it is continuous at \(x = 1\).

Step 4: Consolidate results

Since the function is continuous at every point of its domain, we can conclude that the function \(y\) is continuous for its entire domain which is all real numbers.

Key Concepts

Understanding Piecewise Functions

Piecewise functions are fascinating as they are composed of different expressions, each valid over certain intervals of the function's domain.
They allow us to define functions that behave differently depending on the input value of the variable, typically represented by 'pieces,' each with its own function rule.

• The general format of a piecewise function involves several branches defined with 'if' conditions, specifying the intervals or points they are applicable to.
• It's like constructing different rules or equations for different sections of the x-axis.

In our exercise, the function is defined as:

• For all values of \( x eq 1 \), it is determined by the expression \( \frac{x^{2}-1}{x-1} \).
• At \( x = 1 \), it transitions to a constant value, \( y = 2 \).

Understanding which piece of the expression is used depends on where the input falls, and the transition between these pieces is a crucial aspect of analyzing their continuity and domain.

Exploring the Domain of a Function

The domain of a function refers to the complete set of possible input values (often real numbers) for which the function is defined. Think of it as the collection of all 'x' values you can plug into the function without any issues.

• A key part of working with piecewise functions involves checking the domain of each individual piece to ensure overall cohesiveness.
• Specifically for rational expressions like \( \frac{x^{2}-1}{x-1} \), it’s crucial to identify points where the denominator would be zero, as these points are typically excluded from the domain.

In our exercise, for \( x eq 1 \), \( x - 1 \) must not equal zero, meaning \( x = 1 \) is initially not included in this piece of the domain.
However, since the function is explicitly defined as \( 2 \) at \( x = 1 \), the overall domain becomes all real numbers, ensuring that our function has no gaps anywhere along the x-axis. This is a common feature of many piecewise functions where special cases are addressed separately.

Applying the Continuous Function Theorem

The Continuous Function Theorem provides vital insights into determining the continuity of functions.
According to this theorem, a function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at the point.

• If you have two continuous functions and take their quotient like \( \frac{f(x)}{g(x)} \), it's still continuous wherever the denominator \( g(x) eq 0 \).

By applying this theorem to our piecewise function:

• For \( x eq 1 \), both the numerator \( x^2 - 1 \) and the denominator \( x - 1 \) are polynomials, which are continuous everywhere except at points that might make the denominator zero.
• At \( x = 1 \), the piece is defined as a constant value (\( y = 2 \)), which by its nature is always continuous.

Thus, the function remains continuous across its entire domain. Understanding this theorem helps to seamlessly check continuity for composed expressions defined in pieces, making them analytically robust.