Analyzing F(x) = (x+1)^2 - 1 Inverse And Bijective Properties

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#1. Introduction

In this article, we delve into the fascinating properties of the function f(x) = (x+1)^2 - 1, defined for x ≥ -1. Our main objective is to analyze two crucial statements concerning this function:

Statement 1: The set of solutions to the equation f(x) = f^{-1}(x) is {0, -1}.

Statement 2: The function f is a bijection.

We will rigorously examine each statement, providing detailed explanations and justifications to determine their validity. This exploration will involve understanding the concepts of inverse functions, bijections, and the methods for finding solutions to equations involving functions and their inverses. By the end of this analysis, we will not only ascertain the truthfulness of the statements but also gain a deeper appreciation for the mathematical principles underlying these concepts. This article is designed to provide a comprehensive understanding suitable for students and enthusiasts interested in function analysis and its applications.

#2. Understanding the Function f(x) = (x+1)^2 - 1

Before we dive into the statements, let's first thoroughly understand the function f(x) = (x+1)^2 - 1, defined for x ≥ -1. This restriction on the domain is crucial, as it affects the function's behavior and its invertibility. The function is a quadratic, but the restricted domain transforms it into a function with specific properties that are essential to our analysis.

Key Characteristics of f(x)

  1. Quadratic Nature: The function f(x) is a quadratic function, which means its graph is a parabola. The general form of a quadratic function is ax^2 + bx + c. In our case, expanding f(x) gives us x^2 + 2x, which is a parabola opening upwards.

  2. Vertex and Minimum Value: The vertex of the parabola is a critical point. The x-coordinate of the vertex can be found using the formula -b/2a. For f(x) = x^2 + 2x, this is -2/(21) = -1*. The y-coordinate of the vertex is f(-1) = (-1+1)^2 - 1 = -1. Therefore, the vertex of the parabola is at the point (-1, -1). Since the parabola opens upwards, this is the minimum point of the function.

  3. Domain Restriction: The function is defined for x ≥ -1. This restriction is vital because it ensures that the function is one-to-one, a necessary condition for it to have an inverse. Without this restriction, the parabola would fail the horizontal line test, meaning a horizontal line could intersect the graph at more than one point, indicating that the function is not one-to-one.

  4. Range: Given the domain x ≥ -1 and the vertex at (-1, -1), the range of the function is y ≥ -1. This means that the function's output values are all real numbers greater than or equal to -1. Understanding the range is crucial when we discuss the inverse function, as the range of f(x) becomes the domain of its inverse.

  5. Monotonicity: For x ≥ -1, the function f(x) is monotonically increasing. This means that as x increases, f(x) also increases. This property is essential for the function to be one-to-one and thus invertible over this domain. The increasing nature of the function simplifies the analysis of its inverse and the solutions to equations involving f(x) and its inverse.

By carefully considering these characteristics, we set the stage for a thorough examination of the statements regarding the inverse function and the bijective nature of f(x). Understanding these foundational aspects of the function is key to correctly assessing the given statements.

#3. Statement 1: The Set x f(x) = f^{-1(x)} = {0, -1}

Statement 1 posits that the set of solutions to the equation f(x) = f^{-1}(x) is {0, -1}. This means that the only values of x for which the function f(x) equals its inverse f^{-1}(x) are 0 and -1. To verify this statement, we must first find the inverse function f^{-1}(x) and then solve the equation f(x) = f^{-1}(x). This process involves algebraic manipulation and a careful consideration of the domains and ranges of both f(x) and f^{-1}(x).

Finding the Inverse Function f^{-1}(x)

To find the inverse function, we follow these steps:

  1. Replace f(x) with y: Let y = (x+1)^2 - 1.

  2. Swap x and y: This gives us x = (y+1)^2 - 1.

  3. Solve for y:

    • Add 1 to both sides: x + 1 = (y+1)^2
    • Take the square root of both sides: ±√(x + 1) = y + 1
    • Subtract 1 from both sides: y = -1 ± √(x + 1)

  4. Determine the correct sign: Since the domain of f(x) is x ≥ -1, its range is y ≥ -1. This range becomes the domain of f^{-1}(x), so x ≥ -1 for f^{-1}(x). Also, since f(x) is monotonically increasing for x ≥ -1, its inverse will also be monotonically increasing. This means we should choose the positive square root to ensure the inverse function is also increasing. Thus, f^{-1}(x) = -1 + √(x + 1).

Solving the Equation f(x) = f^{-1}(x)

Now we set f(x) = f^{-1}(x) and solve for x:

(x+1)^2 - 1 = -1 + √(x + 1)

Let u = x + 1. The equation becomes:

u^2 - 1 = -1 + √u u^2 = √u

Square both sides:

u^4 = u u^4 - u = 0 u(u^3 - 1) = 0

This gives us two possible solutions for u:

  1. u = 0
  2. u^3 = 1, which implies u = 1

Now, substitute back x + 1 for u:

  1. x + 1 = 0 implies x = -1
  2. x + 1 = 1 implies x = 0

Therefore, the solutions to f(x) = f^{-1}(x) are x = 0 and x = -1. This confirms that the set of solutions is indeed {0, -1}, making Statement 1 true. This rigorous algebraic solution, combined with our initial understanding of the functions, provides a solid foundation for the correctness of the statement. It showcases the importance of both algebraic manipulation and conceptual understanding in solving such problems.

#4. Statement 2: f is a Bijection

Statement 2 asserts that the function f(x) = (x+1)^2 - 1, defined for x ≥ -1, is a bijection. A bijection, also known as a one-to-one correspondence, is a function that is both injective (one-to-one) and surjective (onto). To determine if f(x) is a bijection, we must verify that it satisfies both these properties. This involves demonstrating that each element in the domain maps to a unique element in the range (injectivity) and that every element in the range has a corresponding element in the domain (surjectivity). Understanding these properties is crucial for determining the overall behavior and invertibility of the function.

Injectivity (One-to-One)

A function is injective if distinct elements in the domain map to distinct elements in the range. In other words, if f(x_1) = f(x_2), then x_1 = x_2. To prove that f(x) is injective, we can use this definition and show that it holds true.

Let's assume f(x_1) = f(x_2) for some x_1, x_2 ≥ -1:

(x_1 + 1)^2 - 1 = (x_2 + 1)^2 - 1

Add 1 to both sides:

(x_1 + 1)^2 = (x_2 + 1)^2

Take the square root of both sides:

√( (x_1 + 1)^2 ) = √( (x_2 + 1)^2 ) |x_1 + 1| = |x_2 + 1|

Since x_1 ≥ -1 and x_2 ≥ -1, both x_1 + 1 and x_2 + 1 are non-negative. Therefore, the absolute value signs can be removed:

x_1 + 1 = x_2 + 1

Subtract 1 from both sides:

x_1 = x_2

Thus, we have shown that if f(x_1) = f(x_2), then x_1 = x_2, which means f(x) is injective. This algebraic proof demonstrates that each input maps to a unique output, a fundamental requirement for injectivity. The careful consideration of the domain restriction is crucial in this proof, as it ensures that the absolute value simplification is valid.

Surjectivity (Onto)

A function is surjective if every element in the range has a corresponding element in the domain. In other words, for every y in the range, there exists an x in the domain such that f(x) = y. The range of f(x) for x ≥ -1 is y ≥ -1. To prove surjectivity, we need to show that for any y ≥ -1, we can find an x ≥ -1 such that f(x) = y.

Let y ≥ -1. We want to find an x ≥ -1 such that:

(x + 1)^2 - 1 = y

Add 1 to both sides:

(x + 1)^2 = y + 1

Take the square root of both sides:

x + 1 = ±√(y + 1)

Subtract 1 from both sides:

x = -1 ± √(y + 1)

Since we require x ≥ -1, we choose the positive square root:

x = -1 + √(y + 1)

Now we need to verify that x ≥ -1 for all y ≥ -1. Since y ≥ -1, y + 1 ≥ 0, and thus √(y + 1) ≥ 0. Therefore, -1 + √(y + 1) ≥ -1, which confirms that x ≥ -1.

We have shown that for any y ≥ -1, there exists an x = -1 + √(y + 1) ≥ -1 such that f(x) = y. This proves that f(x) is surjective. The ability to construct an x for any y in the range is a direct demonstration of surjectivity, highlighting the comprehensive mapping of the domain onto the range.

Conclusion: f is a Bijection

Since f(x) is both injective and surjective, it is a bijection. Therefore, Statement 2 is true. The combination of the proofs for injectivity and surjectivity provides a complete demonstration of the bijective nature of the function. This comprehensive understanding is essential for many advanced mathematical concepts and applications.

#5. Conclusion

In summary, we have rigorously analyzed the function f(x) = (x+1)^2 - 1, defined for x ≥ -1, and evaluated two key statements. Our analysis has shown that:

  1. Statement 1: The set x *f(x) = f^{-1(x)*} = {0, -1} is true. We found the inverse function f^{-1}(x) and solved the equation f(x) = f^{-1}(x), confirming that the solutions are indeed 0 and -1.
  2. Statement 2: f is a bijection is true. We demonstrated that f(x) is both injective (one-to-one) and surjective (onto), thus proving its bijective nature.

Therefore, both statements are true. This comprehensive exploration has not only validated the statements but also deepened our understanding of inverse functions, bijections, and the importance of domain restrictions. This kind of analysis is foundational in mathematics and is crucial for solving more complex problems involving functions and their properties. The combination of algebraic manipulation and conceptual understanding is key to successfully navigating such challenges.