Problem 19
Question
Find the domain and the vertical and horizontal asymptotes (if any). $$f(x)=\frac{x+1}{x^{2}+1}$$
Step-by-Step Solution
Verified Answer
The domain of the function \(f(x) = \frac{x+1}{x^{2}+1}\) is all real numbers. There are no vertical asymptotes. The horizontal asymptote is \(y = 0\).
1Step 1: Determine the Domain of the Function
The domain of a rational function \(f(x) = \frac{p(x)}{q(x)}\) is the set of all real numbers that do not make the denominator equal to zero. In this case, the denominator of the function is \(x^{2}+1\). This equation equals zero when \(x^{2} = -1\), however, there are no real numbers that we can square to get a negative number. Thus, the domain of \(f(x) = \frac{x+1}{x^{2}+1}\) is all real numbers.
2Step 2: Determine the Vertical Asymptotes
The vertical asymptotes of a rational function are the values of \(x\) that make the denominator equal to zero. However, as we found in step 1, there are no real numbers that make the denominator \(x^{2}+1\) equal to zero. Therefore, there are no vertical asymptotes for the function \(f(x) = \frac{x+1}{x^{2}+1}\).
3Step 3: Determine the Horizontal Asymptotes
A rational function has a horizontal asymptote only if the degree of the polynomial in the denominator is greater than or equal to the degree of the polynomial in the numerator. A rational function has a horizontal asymptote \(y = b\) if the coefficients of the highest degree terms in the numerator and denominator are equal, and \(b\) is the ratio of the coefficients of the highest degree terms in the numerator and denominator. In this case, the degree of the polynomial in the denominator (\(x^{2}\)) is greater than the degree of the polynomial in the numerator (\(x\)). Therefore, there is a horizontal asymptote. To find it, we compare the coefficients of the highest degree terms. The coefficient of the \(x\) term in the numerator is 1 and the coefficient of the \(x^{2}\) term in the denominator is also 1. Therefore, the horizontal asymptote is \(y = 0\).
Key Concepts
Domain of a FunctionVertical AsymptotesHorizontal Asymptotes
Domain of a Function
The domain of a function refers to all possible values of the input variable, usually represented as \( x \), for which the function is defined. When dealing with rational functions, these are fractions where one polynomial is divided by another, the domain is determined by the denominator. Specifically, we need to find values of \( x \) that don't make the denominator zero because division by zero is undefined.
In this function, \( f(x) = \frac{x+1}{x^2+1} \), the denominator is \( x^2 + 1 \). To find the domain, check when this denominator equals zero. Solving \( x^2 + 1 = 0 \), we arrive at \( x^2 = -1 \). However, no real number squared gives -1, as squares are always non-negative. Hence, there are no restrictions from the denominator—meaning every real number is acceptable.
In this function, \( f(x) = \frac{x+1}{x^2+1} \), the denominator is \( x^2 + 1 \). To find the domain, check when this denominator equals zero. Solving \( x^2 + 1 = 0 \), we arrive at \( x^2 = -1 \). However, no real number squared gives -1, as squares are always non-negative. Hence, there are no restrictions from the denominator—meaning every real number is acceptable.
- The domain of the function is all real numbers: \( \mathbb{R} \).
Vertical Asymptotes
Vertical asymptotes occur in rational functions where the denominator is zero, but the overall function is undefined. They indicate a value that \( x \) approaches, causing the function to surge towards infinity or negative infinity. Finding vertical asymptotes involves solving for \( x \) where the denominator equals zero.
For \( f(x) = \frac{x+1}{x^2+1} \), we again look at when \( x^2 + 1 = 0 \). As mentioned, there are no real \( x \)-values that satisfy \( x^2 = -1 \).
For \( f(x) = \frac{x+1}{x^2+1} \), we again look at when \( x^2 + 1 = 0 \). As mentioned, there are no real \( x \)-values that satisfy \( x^2 = -1 \).
- Conclusion: This function lacks vertical asymptotes because its denominator never zeros out with real numbers.
Horizontal Asymptotes
Horizontal asymptotes describe the end behavior of a rational function, indicating the value that the function approaches as \( x \) goes to infinity or negative infinity. To find horizontal asymptotes, compare the degrees of the highest terms of the numerator and the denominator.
For \( f(x) = \frac{x+1}{x^2+1} \), the highest degree in the numerator is 1 (from \( x \)) and in the denominator it's 2 (from \( x^2 \)). Since the degree of the denominator exceeds the numerator, the horizontal asymptote is \( y = 0 \) by default. This occurs because, as \( x \) grows indefinitely, the value of \( x^2 \) in the denominator dominates, driving the value of the fraction towards zero.
For \( f(x) = \frac{x+1}{x^2+1} \), the highest degree in the numerator is 1 (from \( x \)) and in the denominator it's 2 (from \( x^2 \)). Since the degree of the denominator exceeds the numerator, the horizontal asymptote is \( y = 0 \) by default. This occurs because, as \( x \) grows indefinitely, the value of \( x^2 \) in the denominator dominates, driving the value of the fraction towards zero.
- Therefore, the horizontal asymptote is at \( y = 0 \).
Other exercises in this chapter
Problem 19
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Determine whether the function is a polynomial function. If so, find the degree. If not, state the reason. $$f(x)=5$$
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