Q.Let $f = \left\{ \left(x,\ \dfrac{x^2}{1 + x^2}\right) : x \in \mathbb{R} \right\}$ be a function from $\mathbb{R}$ into $\mathbb{R}$. Determine the range of $f$.

The key idea is that the range of a rational function can be found by solving for $x$ in terms of $y$ and applying the domain restriction $x \in \mathbb{R}$.

Let $y = \frac{x^2}{1 + x^2}$. Since $x^2 \geq 0$ for all real $x$, the denominator $1 + x^2 > 0$, so $y$ is always defined.

Step 1: Solve for $x^2$.

$y(1 + x^2) = x^2 \implies y + y x^2 = x^2 \implies y = x^2(1 - y)$.

Step 2: Isolate $x^2$.

$x^2 = \frac{y}{1 - y}$, provided $y \neq 1$.

Step 3: For $x$ to be real, $x^2 \geq 0$. So $\frac{y}{1 - y} \geq 0$. …

The core idea here is simple: we have a rational function where both numerator and denominator are always non‑negative. The domain is all real numbers, so the only question is: what possible outputs can this fraction produce?

When you see $\frac{x^2}{1+x^2}$, notice that the denominator is always larger than the numerator (except when $x=0$, they are equal). That immediately tells you the fraction is always less than $1$. And since $x^2 \ge 0$, the fraction is never negative. So the range is squeezed between $0$ and $1$ — but we need to check whether $0$ and $1$ are actually attained, and whether every number in between is hit.

Let’s work through it systematically.

1. Check the lower bound.

   At $x = 0$, we get $f(0) = \frac{0}{1+0} = 0$. So $0$ is in the range.

   For any $x \neq 0$, $x^2 > 0$, so $f(x) > 0$. Thus $0$ is the minimum value.

2. Check the upper bound.

   Can $f(x) = 1$? That would require $\frac{x^2}{1+x^2} = 1$, which gives $x^2 = 1 + x^2$, i.e. $0 = 1$, impossible. So $1$ is never attained.

   But as $|x|$ grows very large, $x^2$ dominates the $1$ in the denominator, so $f(x)$ gets arbitrarily close to $1$. Formally, $\lim_{x \to \pm\infty} \frac{x^2}{1+x^2} = 1$. So $1$ is a supremum (least upper bound) but not a maximum.

3. Do we get every value between $0$ and $1$?

   Let $y$ be any number in $(0,1)$. We want to know if there exists an $x$ such that $\frac{x^2}{1+x^2} = y$.

   Solve for $x^2$: …