Problem 10
Question
For the following exercises, find the domain, vertical asymptotes, and horizontes of the functions. $$ f(x)=\frac{4}{x-1} $$
Step-by-Step Solution
Verified Answer
Domain: \( x \in \mathbb{R}, x \neq 1 \); Vertical asymptote: \( x=1 \); Horizontal asymptote: \( y=0 \).
1Step 1: Find the Domain
To find the domain of the function, we need to identify any restrictions on the input variable, \( x \). The function \( f(x) = \frac{4}{x-1} \) is undefined where the denominator is zero because division by zero is undefined. So we set the denominator equal to zero and solve for \( x \): \[ x-1 = 0 \] \[ x = 1 \] Therefore, \( x = 1 \) is not in the domain. The domain of \( f(x) \) is all real numbers except \( x = 1 \), written as \( x \in \mathbb{R}, x eq 1 \).
2Step 2: Identify the Vertical Asymptotes
A vertical asymptote of a function occurs where the function approaches infinity or negative infinity, often where the function is undefined. For \( f(x) = \frac{4}{x-1} \), vertical asymptotes occur where the denominator equals zero. From the previous step, we found that the function is undefined at \( x = 1 \). This means there is a vertical asymptote at \( x = 1 \).
3Step 3: Identify the Horizontal Asymptotes
Horizontal asymptotes of a function describe the behavior of the function as \( x \) approaches infinity or negative infinity. For rational functions like \( f(x) = \frac{4}{x-1} \), compare the degrees of the numerator and denominator. The numerator has a degree of 0 (since it is a constant \( 4 \)), and the denominator has a degree of 1 (since it is \( x - 1 \)).Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at \( y = 0 \). This means as \( x \to \pm\infty \), \( f(x) \to 0 \).
Key Concepts
Understanding the Domain of Rational FunctionsExploring Vertical AsymptotesUnraveling Horizontal Asymptotes
Understanding the Domain of Rational Functions
The domain of a rational function refers to all the possible values of the variable, typically denoted as \( x \), for which the function is defined. In simpler terms, it represents all the \( x \) values you can plug into the function without causing any mathematical errors.
For the rational function \( f(x) = \frac{4}{x-1} \), we need to focus on the denominator, \( x-1 \), because division by zero is undefined. To find where the function is undefined, solve \( x-1=0 \), which gives \( x=1 \).
This indicates that \( x=1 \) is not included in the domain of \( f(x) \). Therefore, the domain is all real numbers excluding \( x=1 \), expressed as \( x \in \mathbb{R}, \ x eq 1 \).
For the rational function \( f(x) = \frac{4}{x-1} \), we need to focus on the denominator, \( x-1 \), because division by zero is undefined. To find where the function is undefined, solve \( x-1=0 \), which gives \( x=1 \).
This indicates that \( x=1 \) is not included in the domain of \( f(x) \). Therefore, the domain is all real numbers excluding \( x=1 \), expressed as \( x \in \mathbb{R}, \ x eq 1 \).
- Key Point: Domains exclude values where the denominator equals zero.
- Important: Always solve for \( x \) in the denominator to define the domain correctly.
Exploring Vertical Asymptotes
Vertical asymptotes occur at the values of \( x \) that make the denominator of a rational function equal to zero. These are points where the function heads towards positive or negative infinity, indicating a sharp curve on the graph.
In the function \( f(x) = \frac{4}{x-1} \), we've determined that the denominator becomes zero at \( x=1 \). Consequently, a vertical asymptote exists at this point.
A vertical asymptote essentially creates a boundary line on the graph that the function approaches but never crosses.
In the function \( f(x) = \frac{4}{x-1} \), we've determined that the denominator becomes zero at \( x=1 \). Consequently, a vertical asymptote exists at this point.
A vertical asymptote essentially creates a boundary line on the graph that the function approaches but never crosses.
- Vertical asymptotes tell us where the function spikes abruptly.
- These asymptotes occur at values excluded from the domain.
Unraveling Horizontal Asymptotes
Horizontal asymptotes provide insight into the behavior of a function as \( x \) becomes extremely large, either positively or negatively. To determine their presence in a rational function, compare the degrees of the numerator and the denominator.
For \( f(x) = \frac{4}{x-1} \), the numerator is a constant, indicating a degree of zero, while the denominator has a degree of one. When the degree of the denominator is higher than the numerator, \( f(x) \) approaches zero as \( x \to \pm \infty \). This means the horizontal asymptote exists at \( y=0 \).
Horizontal asymptotes guide us on how the function levels off at its tails and provide a sense of stability as \( x \) increases or decreases without bound.
For \( f(x) = \frac{4}{x-1} \), the numerator is a constant, indicating a degree of zero, while the denominator has a degree of one. When the degree of the denominator is higher than the numerator, \( f(x) \) approaches zero as \( x \to \pm \infty \). This means the horizontal asymptote exists at \( y=0 \).
Horizontal asymptotes guide us on how the function levels off at its tails and provide a sense of stability as \( x \) increases or decreases without bound.
- If the numerator's degree is less than the denominator's, expect a horizontal asymptote at \( y=0 \).
- These asymptotes help predict long-term behavior of rational functions.
Other exercises in this chapter
Problem 9
Identify the function as a power function, a polynomial function, or neither. $$f(x)=\frac{x^{2}}{x^{2}-1}$$
View solution Problem 10
For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies inversely as \(x\) and when \(x=4, y=2\).
View solution Problem 10
For the following exercises, find the inverse of the function on the given domain. $$ f(x)=9-x^{2},[0, \infty) $$
View solution Problem 10
For the following exercises, use the Remainder Theorem to find the remainder. $$ \left(5 x^{5}-4 x^{4}+3 x^{3}-2 x^{2}+x-1\right) \div(x+6) $$
View solution