Problem 31
Question
If the point \((a, b)\) lies on the graph of \(f\) and \(f\) has an inverse, then the point ____ lies on the graph of \(f^{-1}\).
Step-by-Step Solution
Verified Answer
The point \((b, a)\) lies on the graph of \(f^{-1}\).
1Step 1: Understanding the Function and Its Inverse
A function, denoted here as \(f\), maps inputs to outputs, such that \(f(a) = b\). If \(f\) has an inverse function, written as \(f^{-1}\), it means that the inverse function will map the output back to the input.
2Step 2: Reversing the Coordinates for the Inverse
For the inverse function \(f^{-1}\), the roles of the input and output are switched. If the point \((a, b)\) lies on the original function \(f\), then the point \((b, a)\) will lie on the inverse function \(f^{-1}\). This is because \(f^{-1}(b) = a\).
Key Concepts
Function MappingCoordinatesGraph Analysis
Function Mapping
In mathematics, a function is fundamentally a rule that assigns each element from one set to exactly one element in another set. We often think of this as sending an input to an output. For example, the function \(f\) can be pictured as a machine that takes an input \(a\) and produces an output \(b\). Thus, we say \(f(a) = b\). If this function has an inverse, denoted as \(f^{-1}\), it can reverse this process, mapping the output back to the input, meaning \(f^{-1}(b) = a\).
When dealing with functions and their inverses, it's crucial to understand that not all functions possess inverses. An inverse function exists only if every input is uniquely paired with an output, emphasizing the importance of one-to-one mappings. In practical terms, for \(f^{-1}\) to be valid, each value \(b\) should be mapped back to a unique \(a\) through \(f^{-1}\).
This process of mapping and reversing is essential in understanding deeper mathematical concepts such as bijections and is often visualized through coordinate changes.
When dealing with functions and their inverses, it's crucial to understand that not all functions possess inverses. An inverse function exists only if every input is uniquely paired with an output, emphasizing the importance of one-to-one mappings. In practical terms, for \(f^{-1}\) to be valid, each value \(b\) should be mapped back to a unique \(a\) through \(f^{-1}\).
This process of mapping and reversing is essential in understanding deeper mathematical concepts such as bijections and is often visualized through coordinate changes.
Coordinates
Coordinates provide a systematic way of locating points on a graph. In the context of functions, coordinates are written as ordered pairs, for example, \((a, b)\). This notation tells us that when we input \(a\) into the function \(f\), we obtain \(b\) as the output.
When working with inverse functions, these coordinates swap their positions. If \((a, b)\) is on the graph of \(f\), the graph of the inverse function \(f^{-1}\) naturally includes the point \((b, a)\). In essence, the input becomes the output, and vice versa. This switch is central when plotting inverse functions and understanding their graphical representation.
Using coordinates, we can easily spot these reflections over the line \(y = x\) in the Cartesian plane. This simple coordinate swap is a powerful tool for dealing with inverse functions and ensures that the essential relationships in function mappings are maintained.
When working with inverse functions, these coordinates swap their positions. If \((a, b)\) is on the graph of \(f\), the graph of the inverse function \(f^{-1}\) naturally includes the point \((b, a)\). In essence, the input becomes the output, and vice versa. This switch is central when plotting inverse functions and understanding their graphical representation.
Using coordinates, we can easily spot these reflections over the line \(y = x\) in the Cartesian plane. This simple coordinate swap is a powerful tool for dealing with inverse functions and ensures that the essential relationships in function mappings are maintained.
Graph Analysis
Graph analysis involves understanding and interpreting the visual representation of functions and their inverses. By analyzing graphs, we can gain insights into the behavior of the function and its inverse. When a point \((a, b)\) lies on the graph of \(f\), the corresponding point \((b, a)\) will lie on the graph of \(f^{-1}\). This illustrates the core concept of inverse functions.
One way to visually verify the relationship between a function and its inverse is to reflect the graph of \(f\) across the line \(y = x\). Any point on \(f\) will have its coordinates flipped in \(f^{-1}\) as a reflection across this line.
Graph analysis is important for identifying where these relationships hold or break down. In more complex functions, analyzing the graph helps determine intervals where the function maintains a one-to-one relationship, and thus, where it can have an inverse. By carefully examining these graphs, one can see the patterns of bijection and the visual symmetry that inverse functions exhibit.
One way to visually verify the relationship between a function and its inverse is to reflect the graph of \(f\) across the line \(y = x\). Any point on \(f\) will have its coordinates flipped in \(f^{-1}\) as a reflection across this line.
Graph analysis is important for identifying where these relationships hold or break down. In more complex functions, analyzing the graph helps determine intervals where the function maintains a one-to-one relationship, and thus, where it can have an inverse. By carefully examining these graphs, one can see the patterns of bijection and the visual symmetry that inverse functions exhibit.
Other exercises in this chapter
Problem 30
Solve each equation. Give the exact answer. $$\log _{1 / 4} \frac{16^{2}}{2^{-3}}=x$$
View solution Problem 31
A construction worker wants to invest \(\$ 60,000\) in a pension plan. One investment offers \(2 \%\) compounded quarterly. Another offers \(1.8 \%\) compounded
View solution Problem 31
Solve each exponential equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the near
View solution Problem 31
Simplify each expression. (a) \(3^{\log _{3} 7}\) (b) \(4^{\log _{4} 9}\) (c) \(12^{\log _{13} 4}\) (d) \(a^{\log _{c} k}(k > 0, a > 0, a \neq 1)\)
View solution