Finding The Inverse Of F(x) = (3x + 5) / (x - 2) A Step-by-Step Guide

Introduction: Understanding Inverse Functions

In mathematics, particularly in the realm of function analysis, the concept of an inverse function plays a crucial role. Before we dive into the specifics of finding the inverse of the function f(x) = (3x + 5) / (x - 2), it's essential to grasp what an inverse function truly represents. Simply put, the inverse function of a given function f(x), often denoted as f⁻¹(x), is a function that "undoes" the action of f(x). In other words, if f(a) = b, then f⁻¹(b) = a. This property highlights the fundamental relationship between a function and its inverse: they essentially reverse each other's mappings. To determine whether a function possesses an inverse function, we employ the horizontal line test. A function has an inverse if and only if no horizontal line intersects its graph more than once. This condition ensures that each output value corresponds to a unique input value, a prerequisite for the existence of an inverse. Moreover, the domain and range of a function and its inverse are intertwined. The domain of f(x) becomes the range of f⁻¹(x), and vice versa. This reciprocal relationship is a cornerstone of understanding inverse functions. Inverse functions have significant applications across various mathematical domains, including calculus, algebra, and analysis. They are instrumental in solving equations, simplifying expressions, and understanding the behavior of functions. In calculus, for instance, inverse functions are used to find derivatives and integrals of complex functions. In algebra, they aid in solving equations involving composite functions. Understanding inverse functions is therefore paramount for any student delving into advanced mathematical concepts.

Step-by-Step Process to Find the Inverse

To find the inverse of the function f(x) = (3x + 5) / (x - 2), we embark on a systematic, multi-step procedure that ensures accuracy and clarity. This process involves a series of algebraic manipulations designed to isolate the dependent variable and express it in terms of the independent variable. First and foremost, we replace f(x) with y. This seemingly simple substitution serves to streamline the subsequent algebraic steps, making the equation easier to manipulate. So, we rewrite the function as y = (3x + 5) / (x - 2). This initial step sets the stage for the core of the inverse function determination process. The next critical step involves swapping x and y. This action embodies the very essence of finding an inverse function – reversing the roles of input and output. By interchanging x and y, we obtain the equation x = (3y + 5) / (y - 2). This equation now represents the inverse relationship, albeit implicitly. Our primary goal now is to explicitly solve this equation for y. To achieve this, we begin by multiplying both sides of the equation by (y - 2). This eliminates the denominator on the right side, simplifying the equation and bringing us closer to isolating y. The resulting equation is x(y - 2) = 3y + 5. This step is pivotal as it transforms the equation into a form that is more amenable to algebraic manipulation. Next, we distribute x on the left side of the equation. This yields xy - 2x = 3y + 5. Expanding the equation in this way allows us to group the terms containing y on one side and the remaining terms on the other. This is a crucial step in isolating y and expressing it as a function of x. Now, we rearrange the equation to group all terms containing y on one side and all other terms on the other side. This involves subtracting 3y from both sides and adding 2x to both sides. The resulting equation is xy - 3y = 2x + 5. This rearrangement brings us closer to the final solution, as it consolidates all the y terms, making them easier to factor out. With the y terms grouped together, we factor out y from the left side of the equation. This gives us y(x - 3) = 2x + 5. Factoring out y is a critical step as it allows us to isolate y by dividing both sides of the equation by the factor (x - 3). Finally, to isolate y, we divide both sides of the equation by (x - 3). This yields the explicit form of the inverse function: y = (2x + 5) / (x - 3). This final step completes the process of finding the inverse, expressing y as a function of x. In the last step, we replace y with f⁻¹(x) to denote the inverse function. Thus, the inverse of f(x) = (3x + 5) / (x - 2) is f⁻¹(x) = (2x + 5) / (x - 3). This notation clearly indicates that we have found the inverse function, which reverses the operation of the original function.

Verifying the Inverse Function

After determining the inverse function, it is imperative to verify its correctness. This verification process ensures that the derived function truly undoes the operation of the original function. The most reliable method to verify an inverse function is to use the composition of functions. Specifically, we need to demonstrate that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. These two conditions must hold true for all x in the appropriate domains for f and f⁻¹. Let’s start by verifying f(f⁻¹(x)) = x. This involves substituting f⁻¹(x) into f(x). Given f(x) = (3x + 5) / (x - 2) and f⁻¹(x) = (2x + 5) / (x - 3), we substitute f⁻¹(x) into f(x) to get:

f(f⁻¹(x)) = f((2x + 5) / (x - 3)) = [3((2x + 5) / (x - 3)) + 5] / [((2x + 5) / (x - 3)) - 2]

Simplifying this expression requires several algebraic steps. First, we find a common denominator in both the numerator and the denominator:

f(f⁻¹(x)) = [(3(2x + 5) + 5(x - 3)) / (x - 3)] / [((2x + 5) - 2(x - 3)) / (x - 3)]

Next, we simplify the numerators:

f(f⁻¹(x)) = [(6x + 15 + 5x - 15) / (x - 3)] / [(2x + 5 - 2x + 6) / (x - 3)]

Further simplification gives us:

f(f⁻¹(x)) = [(11x) / (x - 3)] / [(11) / (x - 3)]

Finally, we divide the fractions, which results in:

f(f⁻¹(x)) = (11x) / 11 = x

Thus, we have verified that f(f⁻¹(x)) = x. Now, we need to verify f⁻¹(f(x)) = x. This involves substituting f(x) into f⁻¹(x):

f⁻¹(f(x)) = f⁻¹((3x + 5) / (x - 2)) = [2((3x + 5) / (x - 2)) + 5] / [((3x + 5) / (x - 2)) - 3]

Again, we simplify this expression by finding a common denominator in both the numerator and the denominator:

f⁻¹(f(x)) = [(2(3x + 5) + 5(x - 2)) / (x - 2)] / [((3x + 5) - 3(x - 2)) / (x - 2)]

Simplifying the numerators, we get:

f⁻¹(f(x)) = [(6x + 10 + 5x - 10) / (x - 2)] / [(3x + 5 - 3x + 6) / (x - 2)]

Further simplification leads to:

f⁻¹(f(x)) = [(11x) / (x - 2)] / [(11) / (x - 2)]

Finally, dividing the fractions gives us:

f⁻¹(f(x)) = (11x) / 11 = x

Hence, we have also verified that f⁻¹(f(x)) = x. Since both conditions, f(f⁻¹(x)) = x and f⁻¹(f(x)) = x, hold true, we can confidently conclude that f⁻¹(x) = (2x + 5) / (x - 3) is indeed the inverse function of f(x) = (3x + 5) / (x - 2). This verification process is a critical step in ensuring the correctness of the derived inverse function and should not be overlooked.

Domain and Range of the Function and Its Inverse

Understanding the domain and range of a function and its inverse is crucial for a comprehensive analysis. The domain of a function is the set of all possible input values (x-values) for which the function is defined, while the range is the set of all possible output values (y-values) that the function can produce. For the original function, f(x) = (3x + 5) / (x - 2), we first determine the domain. The function is a rational function, and rational functions are undefined where the denominator is zero. Therefore, we need to find the values of x for which x - 2 = 0. Solving this equation gives us x = 2. Thus, the domain of f(x) is all real numbers except x = 2, which can be written in interval notation as (-∞, 2) ∪ (2, ∞). To find the range of f(x), we can consider the behavior of the function as x approaches positive and negative infinity, as well as the value that f(x) cannot take. As x becomes very large (positive or negative), the function f(x) approaches the ratio of the leading coefficients, which is 3/1 = 3. This suggests that y = 3 is a horizontal asymptote. To confirm this, we can solve for x in terms of y from the original function: y = (3x + 5) / (x - 2). Multiplying both sides by (x - 2), we get y(x - 2) = 3x + 5. Expanding and rearranging, we have yx - 2y = 3x + 5. Grouping the x terms, we get yx - 3x = 2y + 5, and factoring out x, we have x(y - 3) = 2y + 5. Solving for x, we get x = (2y + 5) / (y - 3). This expression is undefined when y - 3 = 0, which gives us y = 3. Therefore, the range of f(x) is all real numbers except y = 3, which can be written in interval notation as (-∞, 3) ∪ (3, ∞). Now, let's consider the inverse function, f⁻¹(x) = (2x + 5) / (x - 3). The domain of f⁻¹(x) is all real numbers except where the denominator is zero, which occurs when x - 3 = 0. This gives us x = 3. Thus, the domain of f⁻¹(x) is all real numbers except x = 3, or (-∞, 3) ∪ (3, ∞). Notice that the domain of f⁻¹(x) is the range of f(x), which is a characteristic property of inverse functions. To find the range of f⁻¹(x), we again look for horizontal asymptotes. As x becomes very large, f⁻¹(x) approaches the ratio of the leading coefficients, which is 2/1 = 2. To confirm this, we can also look at the expression we derived earlier, x = (2y + 5) / (y - 3), which represents the inverse relationship. Solving this for y gives us the original function, but we already know its domain, which becomes the range of the inverse. The expression f⁻¹(x) = (2x + 5) / (x - 3) is undefined when x = 3, which we already found. Solving for the value that f⁻¹(x) cannot take, we recognize that the range will be all real numbers except 2. Thus, the range of f⁻¹(x) is all real numbers except y = 2, which can be written in interval notation as (-∞, 2) ∪ (2, ∞). Again, we observe that the range of f⁻¹(x) is the domain of f(x), reinforcing the fundamental relationship between a function and its inverse. In summary, for f(x) = (3x + 5) / (x - 2):

• Domain: (-∞, 2) ∪ (2, ∞)
• Range: (-∞, 3) ∪ (3, ∞)

And for f⁻¹(x) = (2x + 5) / (x - 3):

• Domain: (-∞, 3) ∪ (3, ∞)
• Range: (-∞, 2) ∪ (2, ∞)

This detailed analysis of the domain and range provides a complete understanding of the behavior and characteristics of the function and its inverse. It underscores the reciprocal relationship between the two and highlights the importance of these concepts in function analysis.

Conclusion: Significance of Inverse Functions

In conclusion, the process of finding the inverse of a function, such as f(x) = (3x + 5) / (x - 2), involves a methodical approach of interchanging variables and solving for the new dependent variable. This exercise not only provides the inverse function, f⁻¹(x) = (2x + 5) / (x - 3), but also underscores the fundamental concept of inverse relationships in mathematics. The verification step, using the composition of functions, is crucial to ensure the accuracy of the derived inverse. The analysis of the domain and range for both the original function and its inverse further enriches our understanding of their behavior and relationship. Inverse functions are more than just a theoretical construct; they have practical applications across various mathematical fields and real-world scenarios. In calculus, they are essential for finding the derivatives and integrals of functions, especially in complex situations where direct methods may not be feasible. In cryptography, inverse functions play a pivotal role in encryption and decryption processes, ensuring secure communication and data protection. In computer graphics, inverse functions are used for transformations, such as projecting 3D objects onto a 2D screen and vice versa. Understanding inverse functions is also vital in solving equations and modeling real-world phenomena. Many physical and engineering problems involve relationships that can be modeled using functions, and finding the inverse allows us to reverse these relationships, providing valuable insights and solutions. For instance, if a function describes the distance traveled given the time, its inverse would describe the time taken to travel a certain distance. The concept of inverse functions also extends beyond basic algebra and calculus. In more advanced areas of mathematics, such as functional analysis and abstract algebra, inverse functions are generalized to inverse mappings and inverse elements, respectively. These generalizations are fundamental to many theoretical results and applications. Moreover, the study of inverse functions enhances problem-solving skills and mathematical reasoning. The process of finding and verifying an inverse requires a deep understanding of algebraic manipulations, function properties, and the concept of reversibility. It encourages logical thinking and attention to detail, skills that are invaluable in any field. In summary, the ability to find and understand inverse functions is a cornerstone of mathematical literacy. It not only provides a powerful tool for solving mathematical problems but also fosters a deeper appreciation for the interconnectedness of mathematical concepts. From basic algebra to advanced applications, inverse functions play a critical role in both theory and practice.