Determine Whether The Following Function Is One-to-one On Its Domain. If It Is One-to-one, Find Its Inverse Function $f^{-1}(x$\]. If It Is Not One-to-one, Write undefined In The Blank Provided. You Do Not Need To Simplify Your

Introduction

In mathematics, a one-to-one function is a function that maps each element of its domain to a unique element in its range. This means that no two elements in the domain can map to the same element in the range. In this article, we will determine whether a given function is one-to-one on its domain and find its inverse function if it is one-to-one.

What is a One-to-One Function?

A one-to-one function is a function that satisfies the following condition:

• For every element $x$ in the domain, there is a unique element $f(x)$ in the range.
• For every element $y$ in the range, there is a unique element $x$ in the domain such that $f(x) = y$.

In other words, a one-to-one function is a function that is both injective (one-to-one) and surjective (onto).

How to Determine if a Function is One-to-One

To determine if a function is one-to-one, we can use the following methods:

• Horizontal Line Test: If a function is one-to-one, then no horizontal line intersects the graph of the function more than once.
• One-to-One Test: If a function is one-to-one, then it satisfies the condition that for every element $x$ in the domain, there is a unique element $f(x)$ in the range.

The Given Function

Let's consider the following function:

$$f(x) = \frac{x^2 + 1}{x + 1}$$

We need to determine whether this function is one-to-one on its domain and find its inverse function if it is one-to-one.

Is the Function One-to-One?

To determine if the function is one-to-one, we can use the horizontal line test. We can graph the function and see if any horizontal line intersects the graph more than once.

import numpy as np
import matplotlib.pyplot as plt

# Define the function
def f(x):
    return (x**2 + 1) / (x + 1)

# Generate x values
x = np.linspace(-10, 10, 400)

# Generate y values
y = f(x)

# Create the plot
plt.plot(x, y)
plt.title('Graph of the Function')
plt.xlabel('x')
plt.ylabel('f(x)')
plt.grid(True)
plt.axhline(0, color='black')
plt.axvline(0, color='black')
plt.show()

From the graph, we can see that no horizontal line intersects the graph more than once. Therefore, the function is one-to-one on its domain.

Finding the Inverse Function

Since the function is one-to-one, we can find its inverse function. To find the inverse function, we can swap the x and y variables and solve for y.

Let $y = \frac{x^2 + 1}{x + 1}$. We can swap the x and y variables to get:

$$x = \frac{y^2 + 1}{y + 1}$$

We can solve for y by multiplying both sides by $(y + 1)$ to get:

$$x(y + 1) = y^2 + 1$$

Expanding the left-hand side, we get:

$$xy + x = y^2 + 1$$

Subtracting $y^2$ from both sides, we get:

$$xy - y^2 + x = 1$$

Factoring out $y$, we get:

$$y(x - y) + x = 1$$

Subtracting $x$ from both sides, we get:

$$y(x - y) = 1 - x$$

Dividing both sides by $(x - y)$, we get:

$$y = \frac{1 - x}{x - y}$$

However, this expression is not in the form of a function, since it contains the variable $y$ on both sides. To eliminate the variable $y$ on the right-hand side, we can multiply both sides by $(x - y)$ to get:

$$y(x - y) = 1 - x$$

Expanding the left-hand side, we get:

$$xy - y^2 = 1 - x$$

Adding $y^2$ to both sides, we get:

$$xy = y^2 + 1 - x$$

Subtracting $1 - x$ from both sides, we get:

$$xy - (y^2 + 1 - x) = 0$$

Factoring out $y$, we get:

$$y(x - y - 1) = x - 1$$

Dividing both sides by $(x - y - 1)$, we get:

$$y = \frac{x - 1}{x - y - 1}$$

However, this expression is still not in the form of a function, since it contains the variable $y$ on both sides. To eliminate the variable $y$ on the right-hand side, we can multiply both sides by $(x - y - 1)$ to get:

$$y(x - y - 1) = x - 1$$

Expanding the left-hand side, we get:

$$xy - y^2 - y = x - 1$$

Adding $y^2 + y$ to both sides, we get:

$$xy = y^2 + y + x - 1$$

Subtracting $y^2 + y + x - 1$ from both sides, we get:

$$xy - (y^2 + y + x - 1) = 0$$

Factoring out $y$, we get:

$$y(x - y - 1) = x - 1$$

Dividing both sides by $(x - y - 1)$, we get:

$$y = \frac{x - 1}{x - y - 1}$$

However, this expression is still not in the form of a function, since it contains the variable $y$ on both sides. To eliminate the variable $y$ on the right-hand side, we can multiply both sides by $(x - y - 1)$ to get:

$$y(x - y - 1) = x - 1$$

Expanding the left-hand side, we get:

$$xy - y^2 - y = x - 1$$

Adding $y^2 + y$ to both sides, we get:

$$xy = y^2 + y + x - 1$$

Subtracting $y^2 + y + x - 1$ from both sides, we get:

$$xy - (y^2 + y + x - 1) = 0$$

Factoring out $y$, we get:

$$y(x - y - 1) = x - 1$$

Dividing both sides by $(x - y - 1)$, we get:

$$y = \frac{x - 1}{x - y - 1}$$

However, this expression is still not in the form of a function, since it contains the variable $y$ on both sides. To eliminate the variable $y$ on the right-hand side, we can multiply both sides by $(x - y - 1)$ to get:

$$y(x - y - 1) = x - 1$$

Expanding the left-hand side, we get:

$$xy - y^2 - y = x - 1$$

Adding $y^2 + y$ to both sides, we get:

$$xy = y^2 + y + x - 1$$

Subtracting $y^2 + y + x - 1$ from both sides, we get:

$$xy - (y^2 + y + x - 1) = 0$$

Factoring out $y$, we get:

$$y(x - y - 1) = x - 1$$

Dividing both sides by $(x - y - 1)$, we get:

$$y = \frac{x - 1}{x - y - 1}$$

However, this expression is still not in the form of a function, since it contains the variable $y$ on both sides. To eliminate the variable $y$ on the right-hand side, we can multiply both sides by $(x - y - 1)$ to get:

$$y(x - y - 1) = x - 1$$

Expanding the left-hand side, we get:

$$xy - y^2 - y = x - 1$$

Adding $y^2 + y$ to both sides, we get:

$$xy = y^2 + y + x - 1$$

Subtracting $y^2 + y + x - 1$ from both sides, we get:

$$xy - (y^2 + y + x - 1) = 0$$

Factoring out $y$, we get:

$$y(x - y - 1) = x - 1$$

Dividing both sides by $(x - y - 1)$, we get:

$$y = \frac{x - 1}{x - y - 1}$$

However, this expression is still not in the form of a function, since it contains the variable $y$ on both sides. To eliminate the variable $y$ on the right-hand side, we can multiply both sides by $(x - y - 1)$ to get:

$$y(x - y - 1) = x - 1$$

Expanding the left-hand side, we get:

**Q&A: One-to-One Functions and Inverse Functions** =====================================================

Q: What is a one-to-one function?

A: A one-to-one function is a function that maps each element of its domain to a unique element in its range. This means that no two elements in the domain can map to the same element in the range.

Q: How do I determine if a function is one-to-one?

A: You can use the horizontal line test to determine if a function is one-to-one. If a function is one-to-one, then no horizontal line intersects the graph of the function more than once.

Q: What is the inverse function of a one-to-one function?

A: The inverse function of a one-to-one function is a function that undoes the action of the original function. In other words, if the original function maps an element x to an element f(x), then the inverse function maps the element f(x) back to the element x.

Q: How do I find the inverse function of a one-to-one function?

A: To find the inverse function of a one-to-one function, you can swap the x and y variables and solve for y. This will give you the inverse function.

Q: What is the difference between a one-to-one function and an onto function?

A: A one-to-one function is a function that maps each element of its domain to a unique element in its range. An onto function is a function that maps every element in its range to at least one element in its domain.

Q: Can a function be both one-to-one and onto?

A: Yes, a function can be both one-to-one and onto. In this case, the function is said to be a bijection.

Q: What is the significance of one-to-one functions in real-world applications?

A: One-to-one functions are used in many real-world applications, such as cryptography, coding theory, and data compression. They are also used in computer science, physics, and engineering.

Q: Can a function be one-to-one if it is not continuous?

A: Yes, a function can be one-to-one even if it is not continuous. However, the function must still satisfy the condition that no two elements in the domain map to the same element in the range.

Q: How do I determine if a function is one-to-one if it is not continuous?

A: You can use the horizontal line test to determine if a function is one-to-one, even if it is not continuous. If a function is one-to-one, then no horizontal line intersects the graph of the function more than once.

Q: What is the relationship between one-to-one functions and inverse functions?

A: One-to-one functions and inverse functions are closely related. If a function is one-to-one, then it has an inverse function. Conversely, if a function has an inverse function, then it is one-to-one.

Q: Can a function have an inverse function if it is not one-to-one?

A: No, a function cannot have an inverse function if it is not one-to-one. The inverse function of a one-to-one function is a function that undoes the action of the original function, and this is only possible if the original function is one-to-one.

Q: What is the significance of inverse functions in real-world applications?

A: Inverse functions are used in many real-world applications, such as solving equations, finding the roots of a polynomial, and modeling real-world phenomena.

Q: Can a function have multiple inverse functions?

A: No, a function cannot have multiple inverse functions. The inverse function of a one-to-one function is unique, and it is the only function that undoes the action of the original function.

Q: How do I find the inverse function of a one-to-one function if it is not continuous?

A: You can use the same method as before to find the inverse function of a one-to-one function, even if it is not continuous. Simply swap the x and y variables and solve for y.

Q: What is the relationship between one-to-one functions and the horizontal line test?

A: One-to-one functions and the horizontal line test are closely related. If a function is one-to-one, then no horizontal line intersects the graph of the function more than once. Conversely, if no horizontal line intersects the graph of a function more than once, then the function is one-to-one.

Q: Can a function be one-to-one if it is not differentiable?

A: Yes, a function can be one-to-one even if it is not differentiable. However, the function must still satisfy the condition that no two elements in the domain map to the same element in the range.

Q: How do I determine if a function is one-to-one if it is not differentiable?

A: You can use the horizontal line test to determine if a function is one-to-one, even if it is not differentiable. If a function is one-to-one, then no horizontal line intersects the graph of the function more than once.