Problem 72
Question
For each function as defined that is one-to-one, (a) write an equation for the inverse function in the form y = ƒ -11x2, (b) graph ƒ and ƒ -1 on the same axes, and (c) give the domain and the range of ƒ and ƒ -1. If the function is not one-to-one, say so. $$f(x)=\frac{x+2}{x-1}, \quad x \neq 1$$
Step-by-Step Solution
Verified Answer
The inverse function is \( f^{-1}(x) = \frac{x+2}{x-1} \). Both functions are graphed as reflections across y = x. The domain and range for both are all reals except x = 1.
1Step 1: Determine if the function is one-to-one
A function is one-to-one if each value of the output corresponds to exactly one value of the input. To check this, we must see if any horizontal line would intersect the graph of the function more than once. For rational functions like this, it's essential to understand its behavior by finding its derivative and ensuring it does not equal zero for its domain.
2Step 2: Find the inverse of the function
Assuming the function is one-to-one, we need to find the inverse. Start with the equation \[ y = \frac{x+2}{x-1} \]Interchange x and y and solve for y: \[ x = \frac{y+2}{y-1} \]Multiply both sides by \( y-1 \): \[ x(y-1) = y+2 \]Distribute x: \[ xy - x = y + 2 \]Rearrange to solve for y: \[ xy - y = x+2 \]\[ y(x-1) = x+2 \]\[ y = \frac{x+2}{x-1} \]Thus, the inverse function is \[ y = f^{-1}(x) = \frac{x+2}{x-1} \]
3Step 3: Graph the function and its inverse
To graph both functions, note that they will be reflections of each other over the line y = x. Use tools like graphing calculators or plotting software to accurately graph \[ f(x) = \frac{x+2}{x-1} \]and its inverse \[ f^{-1}(x) = \frac{x+2}{x-1} \].
4Step 4: Determine the domain and range
For the original function \( f(x) = \frac{x+2}{x-1} \), the domain is all real numbers except where the denominator is zero, i.e., \( x eq 1 \). The range of \( f(x) \) is all real numbers except where the function is undefined.For the inverse function \( f^{-1}(x) \), the domain and range are the same as \( f(x) \) because \( f(x) \) maps each input to a unique output and vice versa.
Key Concepts
Inverse FunctionsDomain and RangeGraphing Functions
Inverse Functions
Inverse functions provide a way to 'reverse' a function. If you have a function \( f(x) \) that takes some input \( x \) and gives an output \( y \), the inverse function, denoted \( f^{-1}(x) \), works the other way around. It takes \( y \) as input and returns the original \( x \). This is useful for solving equations and understanding relationships between variables.
Consider our example function \( f(x) = \frac{x+2}{x-1} \). To find its inverse, we begin by rewriting \( f(x) \) as \( y = \frac{x+2}{x-1} \). Then, we solve for \( y \) in terms of \( x \):
Therefore, the inverse function is \( f^{-1}(x) = \frac{x+2}{x-1} \). Note that our inverse looks the same as our original function in this instance.
Consider our example function \( f(x) = \frac{x+2}{x-1} \). To find its inverse, we begin by rewriting \( f(x) \) as \( y = \frac{x+2}{x-1} \). Then, we solve for \( y \) in terms of \( x \):
- Switch \( x \) and \( y \): \( x = \frac{y+2}{y-1} \).
- Multiply both sides by \( y-1 \): \( x(y-1) = y+2 \).
- Distribute \( x \): \( xy - x = y + 2 \).
- Rearrange the terms to isolate \( y \): \( y(x-1) = x+2 \).
- Divide by \( x-1 \): \( y = \frac{x+2}{x-1} \).
Therefore, the inverse function is \( f^{-1}(x) = \frac{x+2}{x-1} \). Note that our inverse looks the same as our original function in this instance.
Domain and Range
Understanding the domain and range of a function is crucial in mathematics. The domain of a function is the set of all possible input values (\( x \) values) that the function can accept, while the range is the set of all possible output values (\( y \) values) the function can produce.
For our function \( f(x) = \frac{x+2}{x-1} \), the domain excludes any value that makes the denominator zero. In this case, \( x = 1 \) makes the denominator zero, so:
For the inverse function \( f^{-1}(x) \), the domain and range swap places with the original function. Thus: The domain of \( f^{-1}(x) \) is \( x eq 1 \). The range of \( f^{-1}(x) \) is all real numbers.
For our function \( f(x) = \frac{x+2}{x-1} \), the domain excludes any value that makes the denominator zero. In this case, \( x = 1 \) makes the denominator zero, so:
- The domain of \( f(x) \) is \( x eq 1 \).
- The range of \( f(x) \) is all real numbers.
For the inverse function \( f^{-1}(x) \), the domain and range swap places with the original function. Thus:
Graphing Functions
Graphing functions helps us visualize their behavior. For a function and its inverse, the graphs are mirror images over the line \( y = x \). This means any point \( (a, b) \) on the function \( f(x) \) will also have a corresponding point \( (b, a) \) on the inverse function \( f^{-1}(x) \).
For \( f(x) = \frac{x+2}{x-1} \), graphing it involves plotting several points and drawing a curve through them. Similarly, \( f^{-1}(x) = \frac{x+2}{x-1} \) will look like the original function's reflection over the line \( y = x \).
Using a graphing calculator or software can be particularly helpful:
For \( f(x) = \frac{x+2}{x-1} \), graphing it involves plotting several points and drawing a curve through them. Similarly, \( f^{-1}(x) = \frac{x+2}{x-1} \) will look like the original function's reflection over the line \( y = x \).
Using a graphing calculator or software can be particularly helpful:
- Plot key points for \( f(x) \) and \( f^{-1}(x) \).
- Draw the asymptotes, which are lines the function approaches but never touches. For our function, there's a vertical asymptote at \( x = 1 \).
- Mark the points of intersection and reflection over the line \( y = x \).
Other exercises in this chapter
Problem 72
Solve each equation. $$(3 x+7)^{1 / 3}-(4 x+2)^{1 / 3}=0$$
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Solve each equation for the indicated variable. Assume no denominators are \(0 .\) $$\mathscr{A}=\pi r^{2}, \quad \text { for } r$$
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Find each quotient. Write the answer in standard form \(a+b i .\) $$\frac{1-3 i}{1+i}$$
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Solve each rational inequality. Write each solution set in interval notation. $$\frac{3}{x-6} \leq 2$$
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