Animated Solution for Mathematics - Functions: Let $A=\{x\in R:x\text{ is not a positive integer}\}$. Define a function $f:A\to R$ as $f(x)=\frac{2x}{x-1}$ then $f$ is

Welcome, fellow traveler, to the fascinating world of functions. Today, we are going to dissect a problem that hides a beautiful, intricate structure.

We are looking at the function $f:A\to \mathbb{R}$ defined by:

The domain $A$ is defined as the set of all real numbers except for the positive integers. This constraint is the heartbeat of the problem.

Let us first understand our playground, the domain $A$. We are told $x\in \mathbb{R}$ but $x otin \{1, 2, 3, \dots\}$.

This means our function is defined everywhere on the real number line, except for those specific, discrete points. Imagine walking along the $x$-axis and encountering a series of holes at every positive integer.

This is crucial because it prevents the function from ever hitting certain values. If $x=1$, the denominator becomes zero, which is undefined. By excluding all positive integers, we are essentially carving out a specific path for our function.

Now, let us tackle the question of injectivity. A function is injective (or one-to-one) if every output corresponds to exactly one input.

The most elegant way to check this is to see if the function is strictly monotonic—that is, always increasing or always decreasing. We use the derivative $f^{\prime }(x)$.

Applying the quotient rule to $f(x)=\frac{2x}{x-1}$, we let $u=2x$ and $v=x-1$. The derivative is:

Simplifying this, we get:

Since $(x-1{)}^2$ is always positive for any $x$ in our domain, $f^{\prime }(x)$ is always negative. A function with a strictly negative derivative is strictly decreasing.

This means that for any two distinct $x_1$ and $x_2$ in the domain, $f(x_1)$ will never equal $f(x_2)$. Thus, our function is definitively injective.

Finally, we address surjectivity. A function is surjective (or onto) if its range covers the entire codomain, which in this case is $\mathbb{R}$.

To find the range, we set $y=\frac{2x}{x-1}$ and solve for $x$. Cross-multiplying gives $y(x-1)=2x$, which simplifies to $xy-y=2x$.

Rearranging to isolate $x$, we get $x(y-2)=y$, or:

For $x$ to be a real number, $y$ cannot be $2$. This is our horizontal asymptote.

But there is more! Since $x$ cannot be any positive integer, $y$ cannot be the image of any positive integer. For example, $f(2)=4$, $f(3)=3$, and $f(4)=8/3$.

These values are excluded from the range. Because the range is missing $2$ and an infinite sequence of other values, it cannot be equal to the codomain $\mathbb{R}$. Therefore, the function is not surjective.

We have journeyed through the domain, tested the monotonicity, and explored the range. We found that the function is strictly decreasing, making it injective, but the gaps in the domain prevent it from covering the entire codomain.

The final verdict: the function is injective but not surjective.