Question

If a function \( f: \mathbb{R} \to \{ x \in \mathbb{R} : x \in (-1, 1) \} \) is defined as \( f(x) = \frac{x}{1+|x|}, \ x \in \mathbb{R} \), then prove that f is one-one and onto.

Hint:

An alternative way to prove a differentiable function is one-one is to show its derivative is always positive or always negative. For \(f(x) = \frac{x}{1+|x|}\), you can find that \(f'(x) = \frac{1}{(1+|x|)^2}\) for \(x \neq 0\), which is always positive. A function that is always increasing (or decreasing) must be one-one. This calculus-based approach is often faster than the algebraic case-by-case proof.

Answer 1

Step 1: Understanding the Concept:
To prove that a function is one-one and onto (bijective), we need to satisfy two conditions:
1. One-one (Injective): For any two elements \(x_1, x_2\) in the domain, if \(f(x_1) = f(x_2)\), then it must be that \(x_1 = x_2\). Alternatively, a strictly monotonic function is always one-one.
2. Onto (Surjective): For every element \(y\) in the codomain, there must exist at least one element \(x\) in the domain such that \(f(x) = y\). This means the range of the function must be equal to its codomain.
The given function is \(f(x) = \frac{x}{1+|x|}\), with domain \(\mathbb{R}\) and codomain \((-1, 1)\).
First, we can write f(x) as a piecewise function:
\[ f(x) = \begin{cases} \frac{x}{1+x}, & \text{if } x \ge 0 \\[6pt] \frac{x}{1-x}, & \text{if } x < 0 \end{cases} \] Step 2: Proving f is One-One (Injective):
Let \(x_1, x_2 \in \mathbb{R}\) such that \(f(x_1) = f(x_2)\).
Case 1: \(x_1 \ge 0\) and \(x_2 \ge 0\).
\[ \frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \implies x_1(1+x_2) = x_2(1+x_1) \implies x_1 + x_1x_2 = x_2 + x_2x_1 \implies x_1 = x_2 \] Case 2: \(x_1 < 0\) and \(x_2 < 0\).
\[ \frac{x_1}{1-x_1} = \frac{x_2}{1-x_2} \implies x_1(1-x_2) = x_2(1-x_1) \implies x_1 - x_1x_2 = x_2 - x_2x_1 \implies x_1 = x_2 \] Case 3: \(x_1 \ge 0\) and \(x_2 < 0\).
If \(x_1 \ge 0\), then \(f(x_1) = \frac{x_1}{1+x_1} \ge 0\).
If \(x_2 < 0\), then \(f(x_2) = \frac{x_2}{1-x_2} < 0\).
Therefore, \(f(x_1)\) can never be equal to \(f(x_2)\) in this case (unless both are 0, which means \(x_1 = x_2 = 0\), violating \(x_2 < 0\)).
Since in all possible cases, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function f is one-one.
Step 3: Proving f is Onto (Surjective):
Let \(y\) be an arbitrary element from the codomain \((-1, 1)\). We need to find an \(x \in \mathbb{R}\) such that \(f(x) = y\).
Case 1: \(y \in [0, 1)\).
We expect \(x\) to be non-negative, so we use \(f(x) = \frac{x}{1+x}\).
\[ y = \frac{x}{1+x} \implies y(1+x) = x \implies y + yx = x \implies y = x(1-y) \implies x = \frac{y}{1-y} \] Since \(0 \le y < 1\), the numerator \(y \ge 0\) and the denominator \(1-y > 0\), which means \(x \ge 0\). So, a valid \(x\) exists in the domain.
Case 2: \(y \in (-1, 0)\).
We expect \(x\) to be negative, so we use \(f(x) = \frac{x}{1-x}\).
\[ y = \frac{x}{1-x} \implies y(1-x) = x \implies y - yx = x \implies y = x(1+y) \implies x = \frac{y}{1+y} \] Since \(-1 < y < 0\), the numerator \(y < 0\) and the denominator \(1+y > 0\), which means \(x < 0\). So, a valid \(x\) exists in the domain.
Since for any \(y\) in the codomain \((-1, 1)\), we can find a corresponding \(x\) in the domain \(\mathbb{R}\), the function f is onto.
Step 4: Final Answer
The function f is both one-one and onto, hence it is a bijection from \(\mathbb{R}\) to \((-1, 1)\).