Animated Solution for Mathematics - Functions: Let $f:A\to B$ be a function defined as $f(x)=\frac{x-1}{x-2}$, where $A=R-\{2\}$ and $B=R-\{1\}$. Then f is :-

A Journey into Rational Functions

Welcome, fellow traveler, to the elegant world of functions. Today, we are not just solving an algebra problem; we are dissecting the very anatomy of a mathematical relationship.

We are looking at the function $f(x)=\frac{x-1}{x-2}$. At first glance, it is a simple rational expression, but beneath the surface lies a beautiful structure of symmetry and logic.

To determine if this function is invertible, we must embark on a two-part quest: proving it is injective and proving it is surjective.

The Landscape of the Function

Imagine you are standing on the Cartesian plane, looking at the graph of $f(x)=\frac{x-1}{x-2}$. The first thing you notice is the 'forbidden' territory.

The denominator $x-2$ tells us that $x$ can never be $2$. This creates a vertical asymptote, an invisible wall that the function approaches but never touches.

Similarly, as $x$ grows larger and larger, the constant terms $-1$ and $-2$ become negligible, and the function behaves like $\frac{x}{x}$, which is $1$. This gives us a horizontal asymptote at $y=1$.

These asymptotes define the boundaries of our function's influence, and understanding them is the first step toward mastery.

The Quest for Injectivity

To be invertible, a function must be a bijection—it must be both injective and surjective. Let us tackle injectivity first.

We want to know: does every unique input lead to a unique output? We test this by assuming two different inputs, $x_1$ and $x_2$, produce the same output: $f(x_1)=f(x_2)$.

This leads us to the equation:

Now, we perform the algebraic dance. Cross-multiplying gives us:

Expanding both sides, we get:

Notice the magic? The $x_1{x}_2$ terms and the constant $2$ cancel out perfectly on both sides. We are left with:

Rearranging this, we find $-x_1=-x_2$, which simplifies to $x_1=x_2$.

This is the 'Aha!' moment. Because assuming equal outputs forced the inputs to be identical, we have proven that the function is strictly injective.

The Quest for Surjectivity

Now, we must check if the function is onto, or surjective. Can every $y$ in our codomain be reached by some $x$ in our domain?

We set $y=\frac{x-1}{x-2}$ and solve for $x$. Multiplying by the denominator, we get:

This expands to $xy-2y=x-1$. Grouping the $x$ terms gives:

Factoring out $x$, we arrive at $x(y-1)=2y-1$, and finally:

This expression is valid for all $y eq 1$. Since our codomain $B$ is defined as $\mathbb{R}-\{1\}$, every single value of $y$ has a corresponding $x$. The function is surjective!

The Geometric Harmony

We have proven that $f$ is bijective, and therefore, it is invertible. The inverse function, $f^{-1}(y)=\frac{2y-1}{y-1}$, is not just an algebraic result; it is a geometric reflection.

If you were to draw the line $y=x$ on your graph, the curve of $f(x)$ and the curve of $f^{-1}(x)$ would be perfect mirror images of each other. This symmetry is the hallmark of inverse functions.

You have successfully navigated the logic, conquered the algebra, and visualized the geometry. You have not just solved a problem; you have understood the soul of the function.