Which Statement Could Be Used To Explain Why $f(x) = 2x - 3$ Has An Inverse Relation That Is A Function?A. The Graph Of $f(x$\] Passes The Vertical Line Test.B. $f(x$\] Is A One-to-one Function.C. The Graph Of The Inverse Of

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Introduction

In mathematics, an inverse relation is a relation that undoes the action of another relation. In other words, if we have a relation f(x)f(x), its inverse relation fβˆ’1(x)f^{-1}(x) will take the output of f(x)f(x) and return the original input. However, not all inverse relations are functions. In this article, we will explore the concept of inverse relations and functions, and we will determine which statement could be used to explain why f(x)=2xβˆ’3f(x) = 2x - 3 has an inverse relation that is a function.

What is an Inverse Relation?

An inverse relation is a relation that undoes the action of another relation. In other words, if we have a relation f(x)f(x), its inverse relation fβˆ’1(x)f^{-1}(x) will take the output of f(x)f(x) and return the original input. For example, if we have a relation f(x)=2xβˆ’3f(x) = 2x - 3, its inverse relation fβˆ’1(x)f^{-1}(x) will take the output of f(x)f(x) and return the original input.

What is a Function?

A function is a relation between a set of inputs (called the domain) and a set of possible outputs (called the range). In other words, a function is a relation that assigns to each input exactly one output. For example, the relation f(x)=2xβˆ’3f(x) = 2x - 3 is a function because it assigns to each input exactly one output.

Why is it Important to Determine if an Inverse Relation is a Function?

It is important to determine if an inverse relation is a function because not all inverse relations are functions. In fact, an inverse relation can be a relation that assigns to each input more than one output, or even no output at all. This can lead to confusion and errors in mathematical calculations.

Statement A: The Graph of f(x)f(x) Passes the Vertical Line Test

The vertical line test is a test used to determine if a relation is a function. If a vertical line intersects the graph of a relation at more than one point, then the relation is not a function. However, if a vertical line intersects the graph of a relation at exactly one point, then the relation is a function.

Statement B: f(x)f(x) is a One-to-One Function

A one-to-one function is a function that assigns to each input exactly one output, and no two different inputs have the same output. In other words, a one-to-one function is a function that is both injective and surjective.

Statement C: The Graph of the Inverse of f(x)f(x) is a Function

The graph of the inverse of a relation is a relation that undoes the action of the original relation. In other words, if we have a relation f(x)f(x), its inverse relation fβˆ’1(x)f^{-1}(x) will take the output of f(x)f(x) and return the original input.

Which Statement Could be Used to Explain Why f(x)=2xβˆ’3f(x) = 2x - 3 has an Inverse Relation that is a Function?

After analyzing the three statements, we can conclude that Statement B: f(x)f(x) is a one-to-one function is the correct answer. This is because a one-to-one function is a function that assigns to each input exactly one output, and no two different inputs have the same output. In other words, a one-to-one function is a function that is both injective and surjective.

Conclusion

In conclusion, an inverse relation is a relation that undoes the action of another relation. However, not all inverse relations are functions. In this article, we have determined which statement could be used to explain why f(x)=2xβˆ’3f(x) = 2x - 3 has an inverse relation that is a function. We have concluded that Statement B: f(x)f(x) is a one-to-one function is the correct answer.

Why is f(x)=2xβˆ’3f(x) = 2x - 3 a One-to-One Function?

f(x)=2xβˆ’3f(x) = 2x - 3 is a one-to-one function because it assigns to each input exactly one output, and no two different inputs have the same output. In other words, f(x)=2xβˆ’3f(x) = 2x - 3 is a function that is both injective and surjective.

How to Prove that f(x)=2xβˆ’3f(x) = 2x - 3 is a One-to-One Function

To prove that f(x)=2xβˆ’3f(x) = 2x - 3 is a one-to-one function, we need to show that it is both injective and surjective. To show that it is injective, we need to show that if f(x1)=f(x2)f(x_1) = f(x_2), then x1=x2x_1 = x_2. To show that it is surjective, we need to show that for every output yy, there exists an input xx such that f(x)=yf(x) = y.

Proof that f(x)=2xβˆ’3f(x) = 2x - 3 is Injective

Let x1x_1 and x2x_2 be two inputs such that f(x1)=f(x2)f(x_1) = f(x_2). Then, we have:

f(x1)=f(x2)f(x_1) = f(x_2) 2x1βˆ’3=2x2βˆ’32x_1 - 3 = 2x_2 - 3 2x1=2x22x_1 = 2x_2 x1=x2x_1 = x_2

Therefore, if f(x1)=f(x2)f(x_1) = f(x_2), then x1=x2x_1 = x_2. This shows that f(x)=2xβˆ’3f(x) = 2x - 3 is injective.

Proof that f(x)=2xβˆ’3f(x) = 2x - 3 is Surjective

Let yy be any output. We need to show that there exists an input xx such that f(x)=yf(x) = y. Let x=y+32x = \frac{y + 3}{2}. Then, we have:

f(x)=2xβˆ’3f(x) = 2x - 3 f(x)=2(y+32)βˆ’3f(x) = 2\left(\frac{y + 3}{2}\right) - 3 f(x)=y+3βˆ’3f(x) = y + 3 - 3 f(x)=yf(x) = y

Therefore, for every output yy, there exists an input xx such that f(x)=yf(x) = y. This shows that f(x)=2xβˆ’3f(x) = 2x - 3 is surjective.

Conclusion

In conclusion, we have shown that f(x)=2xβˆ’3f(x) = 2x - 3 is a one-to-one function because it is both injective and surjective. This means that f(x)=2xβˆ’3f(x) = 2x - 3 has an inverse relation that is a function.

Why is it Important to Understand Inverse Relations and Functions?

Understanding inverse relations and functions is important because it helps us to solve mathematical problems and equations. Inverse relations and functions are used in many areas of mathematics, including algebra, geometry, and calculus. By understanding inverse relations and functions, we can solve problems and equations that involve functions and relations.

Conclusion

Q: What is an inverse relation?

A: An inverse relation is a relation that undoes the action of another relation. In other words, if we have a relation f(x)f(x), its inverse relation fβˆ’1(x)f^{-1}(x) will take the output of f(x)f(x) and return the original input.

Q: What is a function?

A: A function is a relation between a set of inputs (called the domain) and a set of possible outputs (called the range). In other words, a function is a relation that assigns to each input exactly one output.

Q: Why is it important to determine if an inverse relation is a function?

A: It is important to determine if an inverse relation is a function because not all inverse relations are functions. In fact, an inverse relation can be a relation that assigns to each input more than one output, or even no output at all. This can lead to confusion and errors in mathematical calculations.

Q: How do I determine if an inverse relation is a function?

A: To determine if an inverse relation is a function, you need to check if it is one-to-one. A one-to-one function is a function that assigns to each input exactly one output, and no two different inputs have the same output.

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

A: A one-to-one function is a function that assigns to each input exactly one output, and no two different inputs have the same output. In other words, a one-to-one function is a function that is both injective and surjective.

Q: How do I prove that a function is one-to-one?

A: To prove that a function is one-to-one, you need to show that it is both injective and surjective. To show that it is injective, you need to show that if f(x1)=f(x2)f(x_1) = f(x_2), then x1=x2x_1 = x_2. To show that it is surjective, you need to show that for every output yy, there exists an input xx such that f(x)=yf(x) = y.

Q: What is the difference between an inverse relation and a function?

A: The main difference between an inverse relation and a function is that an inverse relation can be a relation that assigns to each input more than one output, or even no output at all. A function, on the other hand, is a relation that assigns to each input exactly one output.

Q: Can an inverse relation be a function?

A: Yes, an inverse relation can be a function. In fact, if an inverse relation is one-to-one, then it is a function.

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

A: To find the inverse of a function, you need to swap the x and y values of the function. In other words, if you have a function f(x)=yf(x) = y, then the inverse of the function is fβˆ’1(y)=xf^{-1}(y) = x.

Q: What is the importance of understanding inverse relations and functions?

A: Understanding inverse relations and functions is important because it helps us to solve mathematical problems and equations. Inverse relations and functions are used in many areas of mathematics, including algebra, geometry, and calculus. By understanding inverse relations and functions, we can solve problems and equations that involve functions and relations.

Q: Can you give an example of an inverse relation that is not a function?

A: Yes, here is an example of an inverse relation that is not a function:

Let f(x)={(1,2),(2,3),(3,4)}f(x) = \{ (1, 2), (2, 3), (3, 4) \} be a relation. Then, the inverse of the relation is fβˆ’1(x)={(2,1),(3,2),(4,3)}f^{-1}(x) = \{ (2, 1), (3, 2), (4, 3) \}. However, this inverse relation is not a function because it assigns to each input more than one output.

Q: Can you give an example of an inverse relation that is a function?

A: Yes, here is an example of an inverse relation that is a function:

Let f(x)=2xβˆ’3f(x) = 2x - 3 be a function. Then, the inverse of the function is fβˆ’1(x)=x+32f^{-1}(x) = \frac{x + 3}{2}. This inverse relation is a function because it assigns to each input exactly one output.

Conclusion

In conclusion, we have answered some common questions about inverse relations and functions. We have explained what an inverse relation is, what a function is, and how to determine if an inverse relation is a function. We have also provided examples of inverse relations that are not functions and inverse relations that are functions. By understanding inverse relations and functions, we can solve mathematical problems and equations that involve functions and relations.