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Determine \(f^{-1}\) for (a) \(f: \mathbf{R} \rightarrow \mathbf{R}, f(x)=-x ;\) (b) \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}, f(x, y)=(y, x)\); (c) \(f: \mathbf{R}^{2} \rightarrow\left(\mathbf{R} \times \mathbf{R}^{+}\right), f(x, y)=\left(5 x, e^{y}\right)\).

Short Answer

Expert verified
The inverses of the given functions are \(f^{-1}(x) = -x \), \(f^{-1}(x, y) = (y, x) \) and \(f^{-1}(x, y) = (x/5, \ln y) \), respectively.

Step by step solution

01

Inverse of a Real Function

Given the function \( f: \mathbf{R} \rightarrow \mathbf{R}, f(x)=-x \). To find the inverse we switch x to y and solve for y. Therefore, the equation becomes \( x = -y \) which gives \( y = -x \) when solved for y. So, \(f^{-1}(x) = -x \)
02

Inverse of a Real Function of Two Variables

Given the function \( f: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}, f(x, y)=(y, x) \). To find its inverse, simply exchange x and y to get \( f^{-1}(x, y) = (y, x) \).
03

Inverse of a Function of Two Variables into the Product of Real Numbers and Positive Real Numbers

Given the function \( f: \mathbf{R}^{2} \rightarrow\left(\mathbf{R} \times \mathbf{R}^{+}\right), f(x, y)=\left(5 x, e^{y}\right) \). For the inverse function, we equate this function to \( (x, y) \) and solve for original function variables. Thus we have \[ \begin{cases} 5x_1 = x, \ e^{x_2} = y.\end{cases} \] Solving for \( x_1 \) and \( x_2 \) gives \[ \begin{cases} x_1 = x/5, \ x_2 = \ln y.\end{cases} \] So the inverse function is \( f^{-1}(x, y) = (x/5, \ln y) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Real Function Inverses
Understanding real function inverses is a fundamental concept in mathematics. It involves swapping the input with the output of a function to create a new function. For real-valued functions, this means if you have a function such as \( f(x) = -x \), finding the inverse, denoted as \( f^{-1} \), requires us to switch the function's input \( x \) with its output.

If we express the original function in 'y' terms, \( y = -x \), we then solve for 'x' to find the inverse. Doing so for our example leads us to \( x = -y \), which translates back into function notation as \( f^{-1}(x) = -x \). This process of inverting a function is crucial because it allows us to reverse the action of a function, essentially undoing whatever the function did to its input.

In practical terms, function inverses are used to solve equations, model inverse relationships, and even in calculus, to find the antiderivatives of functions.
Two Variable Function Inverses
When considering functions with two variables, the concept of inverses extends to pairs of inputs and outputs. For instance, if we have a function \( f: \textbf{R}^2 \rightarrow \textbf{R}^2 \), which maps ordered pairs to ordered pairs, such as \( f(x, y) = (y, x) \), finding the inverse requires an exchange of the variables' roles.

To find the inverse, denoted as \( f^{-1} \), we swap the outputs with the inputs. For the example \( f(x, y) = (y, x) \), this action directly shows us that the inverse is \( f^{-1}(x, y) = (y, x) \). This operation can be understood as a reflection across the line where \( x = y \) in a two-dimensional coordinate system.

This kind of function and its inverse are particularly simple because their properties are symmetric. However, when dealing with more complex functions of two variables, finding inverses might involve multiple steps and algebraic manipulations.
Functional Notation and Operations
Functional notation is the way we represent functions and their operations. It provides a clear and consistent method to indicate the input and output relationship. For instance, the notation \( f(x) \) signifies that 'f' is a function acting on the input 'x'. Similarly, \( f^{-1}(x) \) denotes the inverse function that acts on 'x'.

Operations involving functions can include addition, subtraction, multiplication, division, and composition. These operations follow specific rules, such as the distributive and associative properties, which make it easier to handle functions in various mathematical scenarios. Importantly, the goal of using functional notation extends beyond making the math visually manageable — it helps in conceptualizing and processing the relationships patterns within the functions.
Exponential and Logarithmic Functions
Two fundamental types of functions in mathematics are exponential and logarithmic functions. Exponential functions have the form \( f(x) = a^x \), where 'a' is a constant, and 'x' is the exponent. They show how quantities grow or decay at a rate proportional to their current value, which is a common occurrence in fields like finance and physics.

The logarithmic function, denoted as \( f(x) = \log_a(x) \), is the inverse of the exponential function. It answers the question: to what power must we raise 'a' to get 'x'? For example, the inverse function given by \( f^{-1}(x, y) = (x/5, \ln y) \) from the original function \( f(x, y) = (5x, e^y) \) involves the natural logarithm \( \ln \), which is the inverse of the natural exponential function \( e^x \).

These two types of functions are deeply interconnected; understanding one provides insight into the other. Furthermore, logarithms are essential for solving equations involving exponentials, just as exponentials are necessary for modeling scenarios with exponential growth or decay.

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Most popular questions from this chapter

a) In how many ways can \(31,100,905\) be factored into three factors, each greater than 1 , if the order of the factors is not relevant? b) Answer part (a), assuming the order of the three factors is relevant. c) In how many ways can one factor \(31,100,905\) into two or more factors where each factor is greater than 1 and no regard is paid to the order of the factors? d) Answer part (c), assuming the order of the factors is to be taken into consideration.

Use the fact that every polynomial equation having real-number coefficients and odd degree has a real root in order to show that the function \(f: \mathbf{R} \rightarrow \mathbf{R}\), defined by \(f(x)=\) \(x^{5}-2 x^{2}+x\), is an onto function. Is \(f\) one-to-one?

With both of their parents working, Thomas, Stuart, and Craig must handle ten weekly chores among themselves. (a) In how many ways can they divide up the work so that everyone is responsible for at least one chore? (b) In how many ways can the chores be assigned if Thomas, as the eldest, must mow the lawn (one of the ten weekly chores) and no one is allowed to be idle?

If \(f: \mathbf{R} \rightarrow \mathbf{R}\) with \(f(x)=x^{n}\), for which \(n \in \mathbf{Z}^{+}\)is \(f\) invertible?

If \(g: \mathbf{Z}^{+} \rightarrow \mathbf{R}\) and \(c \in \mathbf{R}\), we define the function \(c g: \mathbf{Z}^{+} \rightarrow \mathbf{R}\) by \((c g)(n)=c(g(n))\), for each \(n \in \mathbf{Z}^{+}\). Prove that if \(f, g: \mathbf{Z}^{+} \rightarrow \mathbf{R}\) with \(f \in O(g)\), then \(f \in O(c g)\) for any \(c \in \mathbf{R}, c \neq 0\).

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