Problem 20 - College Algebra and Calculus: An Applied Approach

Question

Describe the interval(s) on which the function is continuous. Explain why the function is continuous on the interval(s). If the function has a discontinuity, identify the conditions of continuity that are not satisfied. \(f(x)=\frac{1}{x^{2}+1}\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{1}{x^{2}+1}\) is continuous on its entire domain which is all real numbers. The function doesn't have any discontinuity, so all conditions of continuity are satisfied for this function.

Step 1: Plot the graph of function

Plot the graph of function \(f(x) = \frac{1}{x^{2}+1}\). As we can see, the graph is a smooth curve without any jumps, holes or asymptotes. Hence, we conclude that the function might be continuous on its entire domain.

Step 2: Find the domain of function

Find the domain of the function. The domain of \(f(x)=\frac{1}{x^{2}+1}\) is all real numbers. This is because the denominator, \(x^{2}+1\), cannot be zero as the square of a any real number is never negative, and adding 1 to it makes no difference in this. So, the function is defined for all real numbers.

Step 3: Verify the continuity of function

Verify the continuity of the function in its entire domain. The limit of \(f(x)\) as \(x\) approaches any real number 'a' is \(\frac{1}{a^{2}+1}\), which equals \(f(a)\). Hence the function is continuous at 'a'. As 'a' was any arbitrary number in the domain of \(f(x)\), it means that the function \(f(x)\) is continuous on its entire domain.

Step 4: Identify the condition of continuity not satisfied if function has a discontinuity

As we concluded that the function \(f(x)\) is continuous on its entire domain, there is no point of discontinuity. Hence, there isn't any condition of continuity that is not satisfied.

Key Concepts

Domain of a function

When we talk about the domain of a function, we’re essentially looking for all possible input values (x-values) for which the function is well-defined. For the function \( f(x) = \frac{1}{x^2 + 1} \), the denominator is \( x^2 + 1 \). Since the square of any real number is always non-negative, \( x^2 + 1 \) is never zero.

This means we can plug any real number into the function and always get a valid output. Thus, the domain of this function is all real numbers. Understanding the domain is crucial because it tells us where to analyze the function for continuity and other characteristics later.

Graph of a function

A graphical representation is an effective tool for visualizing the properties of functions. When we graph \( f(x) = \frac{1}{x^2 + 1} \), we observe a smooth curve without any breaks, jumps, or asymptotes across the entirety of real numbers.

The graph presents a function which behaves nicely without any interruptions, indicating that no x-values lead to discontinuity in the function. This visual tool is crucial as it provides an immediate view of whether a function could be continuous just by checking for any visual disruptions.

Conditions for continuity

For a function to be continuous at a given point \( a \), three conditions must be met:

• The function \( f(x) \) must be defined at \( a \).
• The limit of \( f(x) \) as \( x \) approaches \( a \) must exist.
• The limit of \( f(x) \) as \( x \) approaches \( a \) must equal \( f(a) \).

For the function \( f(x) = \frac{1}{x^2 + 1} \), the limit as \( x \) approaches any real number \( a \) is \( \frac{1}{a^2 + 1} \), which is identical to \( f(a) \).

Since these criteria are satisfied for every real number, \( f(x) \) is continuous across its entire domain. There are no discontinuities because all necessary continuity conditions are fulfilled throughout. Understanding these conditions helps students formally justify why a function behaves continuously rather than just assuming it based on a graph or formula.