Problem 19 - 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. \(f(x)=\frac{x}{x^{2}+1}\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{x}{x^{2} + 1}\) is continuous on all real numbers, because the denominator of function never equals to zero and the limit as \(x\) approaches any real number is equal to the function value at that point. Therefore the function has no points of discontinuity.

Step 1: Finding the Domain of the Function

The first step is to find the domain of the function. As the denominator \(x^{2}+1\) is always greater than zero (since \(x^{2}\) is always greater or equal to zero), the function has a real value for all real numbers. Hence, the domain of this function is all real numbers.

Step 2: Checking Condition for Continuity

A function \(\(f(x)\)\) is continuous at a point 'c' if \(lim_{x\to c^-} f(x) = f(c) = lim_{x\to c^+} f(x)\). As the function has a well-defined value at all real numbers (from step 1), we can evaluate the limit at any point 'c'. By substituting 'c' in the function, we find that \(lim_{x\to c^-} f(x) = f(c) = lim_{x\to c^+} f(x)\). This confirms the function maintains its continuity across its entire domain.

Step 3: Identifying Discontinuities

Discontinuity in a function occurs if the conditions of continuity are violated. While evaluating our function \(f(x)\), we encountered no values of 'x' for which the output of \(f(x)\) will be undefined or infinite. Hence, the conclusion is that our function \(f(x)\) does not have any discontinuity.

Key Concepts

Domain of a function

The domain of a function represents all the possible input values (usually 'x' values) for which the function produces real and valid outputs. For the function \(f(x) = \frac{x}{x^2+1}\), let's examine what values of 'x' we can use safely. The denominator \(x^2 + 1\) governs the domain here.

Since the expression \(x^2+1\) is never zero (as \(x^2\) is always non-negative and at its smallest is zero, which quickly becomes 1 when adding 1), there are no restrictions due to division by zero. This means the function can accept all real numbers as inputs.

• To summarize, the domain of this function is all real numbers, represented as \(\mathbb{R}\).

Limits

In mathematics, limits help us analyze how a function behaves as it approaches a particular point. For the continuity of a function, limits are crucial. The function \(f(x) = \frac{x}{x^2+1}\) is a candidate for continuity, so we must check its limit conditions.

Continuity at a point 'c' requires that:

• The limit from the left, \(\lim_{{x \to c^-}} f(x)\), exists.
• The limit from the right, \(\lim_{{x \to c^+}} f(x)\), exists.
• Both limits are equal to \(f(c)\).

This function has limits existing at every real number 'c' because \(x^2+1\) is always non-zero, meaning that as 'x' approaches any value, the function approaches \(f(c)\) smoothly.

Without any asymptotes or jumps, the limits are comfortably satisfied, affirming that this function is continuous everywhere in the domain.

Discontinuities

Discontinuities occur in a function when there is a point or points where the function fails at least one of the conditions of continuity.

For a function to have a discontinuity, it usually experiences one of the following situations:

• An infinite limit (an undefined jack-up)
• A jump or gap in the graph of the function
• A point where the function doesn't exist

In the case of \(f(x) = \frac{x}{x^2+1}\), we do not encounter any of these issues. The function exists and is calculable for every real number without tending to infinity or having any sudden jumps.

This careful design of the function's form tells us it is free of any discontinuities through its domain, which means it is smoothly continuous everywhere.