A function $f: \mathbb{R} \rightarrow \{x \in \mathbb{R}: -1 < x < 1\}$ is defined as $f(x) = \frac{x}{1+|x|}$, then $f$ is:

1. Check for One-One (Injective):

Since the function contains $|x|$, let's analyze it for two cases:

• Case 1: $x \ge 0$

$f(x) = \frac{x}{1+x}$

Differentiating with respect to $x$:

$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$

• Case 2: $x < 0$

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

Differentiating with respect to $x$:

$f'(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} > 0$

Since $f'(x) > 0$ for all $x \in \mathbb{R}$ (except at $x=0$ where the derivative from both sides is $1$), the function is strictly increasing across its entire domain. A strictly monotonic function is always one-one.

2. Check for Onto (Surjective):

We need to see if the range of the function matches its codomain, which is $(-1, 1)$.

• As $x \to \infty$:

$\lim_{x \to \infty} \frac{x}{1+x} = \lim_{x \to \infty} \frac{1}{\frac{1}{x} + 1} = 1$

• As $x \to -\infty$:

$\lim_{x \to -\infty} \frac{x}{1-x} = \lim_{x \to -\infty} \frac{1}{\frac{1}{x} - 1} = -1$

Since the function is continuous and strictly increases from $-1$ to $1$, its range is exactly the open interval $(-1, 1)$.

Because $\text{Range} = \text{Codomain}$, the function is onto.