Proof That The Inverse Function Composition Is An Identity Function

In mathematics, identity functions play a crucial role, acting as the neutral element in function composition. They leave their input unchanged, essentially acting as a mathematical mirror. This article delves into proving that the composite function $f^{-1}(f(x))$ is an identity function when $f(x)$ is defined as $\frac{3x-1}{x+2}$. This involves understanding the concept of inverse functions, function composition, and the properties that define an identity function. We will meticulously walk through the steps required to demonstrate this fundamental concept, providing a clear and comprehensive explanation for readers of all backgrounds.

Understanding Inverse Functions

Before diving into the proof, it's essential to grasp the concept of an inverse function. An inverse function, denoted as $f^{-1}(x)$, essentially "undoes" the operation of the original function, $f(x)$. In simpler terms, if $f(a) = b$, then $f^{-1}(b) = a$. However, not all functions have inverses. For a function to possess an inverse, it must be bijective, meaning it is both injective (one-to-one) and surjective (onto). Injectivity ensures that each element in the range corresponds to a unique element in the domain, while surjectivity ensures that every element in the codomain is mapped to by at least one element in the domain. The function $f(x) = \frac{3x-1}{x+2}$ satisfies these conditions within its domain, which excludes $x = -2$ since that would result in division by zero. To find the inverse of $f(x)$, we typically swap $x$ and $y$ in the equation $y = \frac{3x-1}{x+2}$ and then solve for $y$. This process will be crucial in the subsequent steps of our proof. Understanding these foundational principles of inverse functions is paramount to comprehending why the composite function $f^{-1}(f(x))$ results in an identity function.

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

To prove that $f^{-1}(f(x))$ is an identity function, the first crucial step is to determine the inverse function, $f^{-1}(x)$, given that $f(x) = \frac{3x-1}{x+2}$. As mentioned earlier, the process involves swapping $x$ and $y$ in the equation $y = \frac{3x-1}{x+2}$ and then solving for $y$. Swapping $x$ and $y$ gives us $x = \frac{3y-1}{y+2}$. Now, our goal is to isolate $y$ on one side of the equation. Multiplying both sides by $(y+2)$ yields $x(y+2) = 3y - 1$. Expanding the left side, we get $xy + 2x = 3y - 1$. To group the terms containing $y$, we rearrange the equation to $xy - 3y = -2x - 1$. Factoring out $y$ from the left side gives us $y(x - 3) = -2x - 1$. Finally, dividing both sides by $(x-3)$, we arrive at the inverse function: $y = \frac{-2x-1}{x-3}$. Therefore, $f^{-1}(x) = \frac{-2x-1}{x-3}$. It is important to note that the domain of $f^{-1}(x)$ is all real numbers except for $x = 3$, as this would result in division by zero. Having successfully derived the inverse function, we are now equipped to proceed with the next stage of the proof, which involves composing $f^{-1}(x)$ with $f(x)$. This will demonstrate how the inverse function effectively "undoes" the original function, leading us to the identity function.

Composition of Functions: $f^{-1}(f(x))$

Now that we have both $f(x) = \frac{3x-1}{x+2}$ and its inverse $f^{-1}(x) = \frac{-2x-1}{x-3}$, we can proceed to the heart of the proof: demonstrating that the composition $f^{-1}(f(x))$ results in the identity function. Function composition involves substituting one function into another. In this case, we need to substitute $f(x)$ into $f^{-1}(x)$. This means replacing every instance of $x$ in the expression for $f^{-1}(x)$ with the entire expression for $f(x)$. So, we have:

$f^{-1}(f(x)) = f^{-1}(\frac{3x-1}{x+2}) = \frac{-2(\frac{3x-1}{x+2})-1}{(\frac{3x-1}{x+2})-3}$

This expression might look daunting, but we will simplify it step-by-step. The key is to perform the algebraic manipulations carefully and methodically. First, we need to eliminate the nested fractions in both the numerator and the denominator. This involves multiplying both the numerator and the denominator of the entire expression by the common denominator of the inner fractions, which is $(x+2)$. This will clear the fractions within the fractions, making the expression easier to manage. This process of simplification is crucial for revealing the underlying structure of the composite function and ultimately demonstrating its identity function nature. Once the simplification is complete, we should see that the resulting expression is simply $x$, thus proving that $f^{-1}(f(x))$ is indeed an identity function.

Simplifying the Composite Function

The crucial step in proving $f^{-1}(f(x))$ is an identity function lies in the careful simplification of the composite function we derived in the previous section. Recall that we have:

$f^{-1}(f(x)) = \frac{-2(\frac{3x-1}{x+2})-1}{(\frac{3x-1}{x+2})-3}$

To simplify this expression, we multiply both the numerator and the denominator by $(x+2)$. This yields:

$f^{-1}(f(x)) = \frac{(x+2)[-2(\frac{3x-1}{x+2})-1]}{(x+2)[(\frac{3x-1}{x+2})-3]}$

Distributing $(x+2)$ in both the numerator and the denominator, we get:

$f^{-1}(f(x)) = \frac{-2(3x-1) - (x+2)}{3x-1 - 3(x+2)}$

Now, we expand the terms:

$f^{-1}(f(x)) = \frac{-6x + 2 - x - 2}{3x - 1 - 3x - 6}$

Combining like terms in both the numerator and the denominator, we have:

$f^{-1}(f(x)) = \frac{-7x}{-7}$

Finally, we can simplify the fraction by canceling the common factor of $-7$:

$f^{-1}(f(x)) = x$

This final result demonstrates that the composite function $f^{-1}(f(x))$ simplifies to $x$. This is the defining characteristic of an identity function. Therefore, we have successfully proven that $f^{-1}(f(x))$ is indeed an identity function for the given function $f(x) = \frac{3x-1}{x+2}$.

Conclusion: $f^{-1}(f(x))$ is an Identity Function

In conclusion, we have meticulously demonstrated that if $f(x) = \frac{3x-1}{x+2}$, then $f^{-1}(f(x))$ is an identity function. This was achieved through a step-by-step process that involved first finding the inverse function, $f^{-1}(x)$, by swapping variables and solving for $y$. We then performed function composition, substituting $f(x)$ into $f^{-1}(x)$ to obtain $f^{-1}(f(x))$. The resulting complex expression was then carefully simplified through algebraic manipulation. Multiplying both the numerator and denominator by $(x+2)$, expanding terms, and combining like terms ultimately led us to the result $f^{-1}(f(x)) = x$. This result is the very definition of an identity function, which maps every input to itself. This proof highlights the fundamental relationship between a function and its inverse and underscores the properties of function composition. The identity function plays a crucial role in various areas of mathematics, serving as the neutral element for function composition. Understanding this concept is essential for further studies in algebra, calculus, and other advanced mathematical fields. The rigorous steps taken in this proof provide a solid foundation for understanding and applying the concept of inverse functions and identity functions in more complex scenarios.