Animated Solution for Mathematics - Functions: Let $f:\mathbb{R}\to \mathbb{R}$ be a function such that $f(x)=\frac{x^2+2x+1}{x^2+1}$. Then

Welcome, fellow traveler in the world of mathematics! Today, we are going to dissect a function that might look intimidating at first glance, but hides a beautiful, symmetric structure. We are looking at the function:

Our mission is to determine where this function is 'one-one' (injective) and where it is 'many-one'. Imagine the function as a machine: you feed it an $x$, and it spits out a $y$. If every $y$ comes from exactly one $x$, the machine is one-one. If the machine ever 'turns around'—like a car driving up a hill and then down—it will inevitably hit the same altitude twice, making it many-one.

Before we dive into the heavy calculus, let's simplify our life. The expression $f(x)=\frac{x^2+2x+1}{x^2+1}$ looks like a classic candidate for the quotient rule, but that is a trap! Let's be clever.

Notice the numerator: $x^2+2x+1$. We can rewrite this as $(x^2+1)+2x$. Now, watch the magic happen:

This is much friendlier! We have transformed a complex rational function into a simple constant plus a manageable fraction.

To find out if the function is one-one, we need to know if it is strictly monotonic—meaning it only goes up or only goes down. The tool for this is the derivative, $f^{\prime }(x)$. Let's differentiate our simplified form:

The derivative of the constant $1$ is $0$. For the second term, we apply the quotient rule: $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$. Here, $u=2x$ and $v=x^2+1$. So, $u^{\prime }=2$ and $v^{\prime }=2x$.

Plugging these in, we get:

Simplifying the numerator: $2x^2+2-4x^2=2-2x^2=2(1-x^2)$. Thus, our derivative is:

Now, we look for the critical points where $f^{\prime }(x)=0$. Since the denominator $(x^2+1{)}^2$ is always positive, the derivative is zero only when the numerator is zero: $1-x^2=0$, which gives us $x=1$ and $x=-1$.

These are our turning points. Let's analyze the intervals:

1. In $(-\infty ,-1)$, pick $x=-2$. Then $1-(-2{)}^2=1-4=-3$, which is negative. The function is decreasing.

2. In $(-1,1)$, pick $x=0$. Then $1-0^2=1$, which is positive. The function is increasing.

3. In $(1,\infty )$, pick $x=2$. Then $1-2^2=1-4=-3$, which is negative. The function is decreasing.

Because the function decreases, then increases, then decreases again, it fails the horizontal line test over the entire real line.

However, if we restrict our view to the interval $[1,\infty )$, the function is strictly decreasing. It never turns back. Therefore, it is one-one in $[1,\infty )$, but many-one over the entire domain.

You have just mastered the art of analyzing function behavior! Keep this intuition, and no function will ever be able to hide its secrets from you.