Problem 55
Question
(a) find \(f^{-1}\) and (b) graph \(f\) and \(f^{-1}\) on the same set of axes. $$ f(x)=\frac{2}{x-1} \quad \text { for } x>1 $$
Step-by-Step Solution
Verified Answer
The inverse function is \( f^{-1}(x) = \frac{x+2}{x} \) for \( x > 0 \).
1Step 1: Understand the Inverse Function
An inverse function, denoted as \( f^{-1} \), is a function that reverses the effect of the original function \( f \). If \( f(x)=y \), then \( f^{-1}(y)=x \). Therefore, finding the inverse involves swapping the dependent and independent variables and solving for the independent variable.
2Step 2: Swap and Set Equation
Start with the equation for \( f(x) \): \( y = \frac{2}{x - 1} \). Next, swap \( x \) and \( y \), giving us \( x = \frac{2}{y - 1} \). This prepares us to solve for \( y \).
3Step 3: Solve for the Inverse Function
To find \( y \), first multiply both sides by \( y - 1 \): \( x(y - 1) = 2 \). Simplify the expression: \( xy - x = 2 \). Add \( x \) to both sides to get: \( xy = x + 2 \). Then divide each term by \( x \): \( y = \frac{x + 2}{x} \). Thus, the inverse function is \( f^{-1}(x) = \frac{x + 2}{x} \) for \( x > 0 \).
4Step 4: Graph the Original Function
Graph \( f(x) = \frac{2}{x-1} \) by recognizing that it is a rational function with a vertical asymptote at \( x = 1 \) and a horizontal asymptote at \( y = 0 \). The function is defined for \( x > 1 \) and approaches the asymptotes without crossing them.
5Step 5: Graph the Inverse Function
Graph the inverse function \( f^{-1}(x) = \frac{x+2}{x} \). Recognize that it also has a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 1 \). This function is defined for \( x > 0 \). The graphs of \( f \) and \( f^{-1} \) should reflect across the line \( y = x \).
6Step 6: Verify Graphically
Make sure both functions \( f(x) \) and \( f^{-1}(x) \) are symmetric with respect to the line \( y = x \) in the graph. This symmetry is a crucial feature of functions and their inverses, verifying correctness.
Key Concepts
Graphing FunctionsRational FunctionsAsymptotes
Graphing Functions
Graphing functions is an essential skill in mathematics, as it helps us visualize how a function behaves. When graphing any function, always start by identifying key features like intercepts, asymptotes, and the domain of the function.
For the function given, \[ f(x) = \frac{2}{x - 1} \],we first notice that the function is only defined for \( x > 1 \) because \( x = 1 \) will make the denominator zero, leading to undefined behavior.
For the function given, \[ f(x) = \frac{2}{x - 1} \],we first notice that the function is only defined for \( x > 1 \) because \( x = 1 \) will make the denominator zero, leading to undefined behavior.
- **Domain:** \( x > 1 \)
- **Vertical Asymptote:** \( x = 1 \)
- **Horizontal Asymptote:** Approaches \( y = 0 \) as \( x \) goes to infinity.
Rational Functions
Rational functions are composed of a ratio of two polynomials. They often have intriguing behaviors because of the relationships between their numerator and denominator.
In our example, \[ f(x) = \frac{2}{x - 1} \],we see that it is a simple rational function with the numerator as a constant \( 2 \) and the denominator \( x-1 \). This function showcases the essential property of rational functions having asymptotes where the denominator equals zero, which heavily influences the function's behavior
The inverse function \[ f^{-1}(x) = \frac{x + 2}{x} \],is another rational function with different components creating new asymptotes.
In our example, \[ f(x) = \frac{2}{x - 1} \],we see that it is a simple rational function with the numerator as a constant \( 2 \) and the denominator \( x-1 \). This function showcases the essential property of rational functions having asymptotes where the denominator equals zero, which heavily influences the function's behavior
The inverse function \[ f^{-1}(x) = \frac{x + 2}{x} \],is another rational function with different components creating new asymptotes.
- This function has a vertical asymptote at \( x = 0 \) since the denominator would make the function undefined.
- It also has a horizontal asymptote at \( y = 1 \), showcasing a shift compared to the original function.
Asymptotes
Asymptotes are lines a graph approaches but never actually reaches. They provide a framework for understanding the boundaries and behavior of a function as the input values become very large or very small.
In our exercise, both functions \( f(x) = \frac{2}{x-1} \) and \( f^{-1}(x) = \frac{x+2}{x} \) have asymptotes that help define their behavior.
In our exercise, both functions \( f(x) = \frac{2}{x-1} \) and \( f^{-1}(x) = \frac{x+2}{x} \) have asymptotes that help define their behavior.
- For \( f(x) \), there is a vertical asymptote at \( x = 1 \), since the denominator would become zero here.
- A horizontal asymptote is observed at \( y = 0 \), meaning as \( x \) goes to infinity, the function value approaches zero.
- For the inverse function \( f^{-1}(x) \), the vertical asymptote is at \( x = 0 \) and a horizontal asymptote at \( y = 1 \).
Other exercises in this chapter
Problem 55
Perform the following calculations and express answers to the nearest hundredth. $$ \frac{\ln 2}{\ln 7} $$
View solution Problem 55
Given that \(\log _{2} 5=2.3219\) and \(\log _{2} 7=2.8074\), evaluate each expression by using Properties \(10.5-10.7\) \(\log _{2} \sqrt{7}\)
View solution Problem 55
Graph each of the functions. $$ f(x)=\frac{e^{x}-e^{-x}}{2} $$
View solution Problem 55
What is the solution for \(3^{x}=5\) ? Do you agree that it is between 1 and 2 because \(3^{1}=3\) and \(3^{2}=9\) ? Now graph \(f(x)=3^{x}-5\) and use the ZOOM
View solution