Dive into the fascinating world of surjective functions! We’re tackling a specific type of equation involving surjective functions f(x) for positive real numbers. This problem challenges us to find all possible functions that map positive real numbers to positive real numbers, while satisfying a particular functional equation. Understanding the nature of surjective functions f(x) is key to solving this puzzle.
This exploration delves into a functional equation with surjective functions f(x). We’ll meticulously analyze the equation, looking for patterns and relationships between the input and output values. Our goal is to pinpoint the specific surjective functions f(x) that fit the criteria, and we’ll use various techniques to achieve this. Crucially, we’ll need to consider the constraints imposed by the domain and range, which are positive real numbers.
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Solving a Functional Equation
Problem Statement
Determine all surjective functions ##f: \\mathbb{R}^+ \\to \\mathbb{R}^+## such that for all ##x \\in \\mathbb{R}^+##:
### 2xf(f(x)) = f(x)(x + f(f(x))) ###Solution
Understanding the Problem
The problem asks us to find all functions ##f## that map positive real numbers to positive real numbers and satisfy the given functional equation. The equation involves ##f(x)## and ##f(f(x))##, suggesting a recursive relationship. The surjectivity condition is crucial.
Solving the Problem
Step 1: Simplifying the Equation
Rearranging the given equation, we get:
### 2xf(f(x)) = xf(x) + f(x)f(f(x)) ### ### xf(x) = f(x)f(f(x)) – 2xf(f(x)) ### ### xf(x) = f(x)(f(f(x)) – 2x) ###Step 2: Implication of Injectivity
Dividing both sides by ##f(x)## (since ##f(x) > 0## for all ##x > 0##), we get:
### x = f(f(x)) – 2x ### ### f(f(x)) = 3x ###This implies that ##f## is injective. If ##f## is surjective, then ##f## is bijective, so ##f^{-1}## exists.
Step 3: Using the Inverse Function
Substituting ##y = f(x)##, we have ##f(y) = 3f^{-1}(y)##. Substituting ##x = f^{-1}(y)## gives ##f(f^{-1}(y)) = 3f^{-1}(f^{-1}(y))##. Since ##f(f^{-1}(y)) = y##, we get:
### y = 3f^{-1}(y) ###This implies that ##f^{-1}(y) = \frac{1}{3}y##. Therefore, ##f(x) = 3x##.
Step 4: Verification
Substituting ##f(x) = 3x## into the original equation:
### 2x(f(3x)) = f(x)(x + f(f(x))) ### ### 2x(3(3x)) = 3x(x + 3(3x)) ### ### 18x^2 = 3x(x + 9x) = 3x(10x) = 30x^2 ###This is not true for all ##x > 0##.
Step 5: Correcting the Error
The error was in assuming that ##f(f(x)) = 3x##. The correct approach is to realize that ##f(f(x)) = 3x## implies ##f(x) = x## for all ##x##. This is a contradiction with the original equation. Thus, ##f(x) = x## for all ##x > 0## is the only solution.
Final Solution
The only surjective function that satisfies the given equation is ##f(x) = x## for all ##x \\in \\mathbb{R}^+##.
Step | Equation/Concept | Explanation/Result |
---|---|---|
Problem Statement | Find all surjective functions ##f: \\mathbb{R}^+ \\to \\mathbb{R}^+## such that ### 2xf(f(x)) = f(x)(x + f(f(x))) ### for all ##x \\in \\mathbb{R}^+## | Determine the function ##f## satisfying the given equation and the surjectivity condition. |
Simplification | ### xf(x) = f(x)(f(f(x)) – 2x) ### ### f(f(x)) = 3x ### | Simplifying the given equation to isolate ##f(f(x))##. |
Injectivity Implication | ##f## is injective. | Dividing by ##f(x)## (since ##f(x) > 0##) implies ##f(f(x)) = 3x##, suggesting injectivity of ##f##. |
Inverse Function | ##f^{-1}(y) = \frac{1}{3}y## ##f(x) = 3x## | Using the inverse function ##f^{-1}## and the relationship ##f(f^{-1}(y)) = y##. |
Verification (Incorrect) | ### 18x^2 = 30x^2 ### | Substituting ##f(x) = 3x## into the original equation, leading to a contradiction. |
Error Correction | ##f(x) = x## | Realizing the error in the previous step and concluding that the only solution is ##f(x) = x##. |
Final Solution | ##f(x) = x## for all ##x \\in \\mathbb{R}^+## | The surjective function satisfying the given equation is ##f(x) = x##. |
The solution demonstrates a crucial step in understanding surjective functions f(x). The initial attempt to find a solution using the inverse function, while seemingly logical, ultimately led to a contradiction. This highlights the importance of carefully considering the properties of surjective functions, particularly their relationship to their inverses, and how those properties interact with the functional equation.
The solution process also underscores the significance of verifying the proposed solution. The initial attempt, though flawed, served to illustrate the importance of careful consideration and rigorous testing. Substituting the proposed solution back into the original equation allowed us to identify the error in the initial approach. This verification step is essential in problem-solving, particularly when dealing with complex functional equations involving surjective functions f(x).
- Importance of Verification: Always verify the solution by substituting it back into the original equation to ensure it satisfies all the conditions.
- Surjectivity and Injectivity: The surjectivity of the function plays a crucial role in the solution. A bijective function is both injective (one-to-one) and surjective (onto). The interplay between these properties is essential to understand.
- Recursive Nature of the Equation: The equation involves f(f(x)), which introduces a recursive relationship. Understanding how to handle such relationships is key to solving these types of problems.
- Domain and Range Constraints: The problem specifies that the domain and range are positive real numbers. These constraints significantly affect the possible solutions.
- Domain and Range Constraints: The problem specifies that the domain and range are positive real numbers. These constraints significantly affect the possible solutions.
This example reinforces the importance of a systematic approach to problem-solving, particularly when dealing with functional equations involving surjective functions f(x). It highlights the need for careful consideration of all possible solutions and the crucial role of verification in confirming the validity of a solution.
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- Alpha Examples: Injectivity & Surjectivity
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