A function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{x^2+1}$ is (where $\mathbb{R}$ is a set of real number)

A function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{x^2+1}$ needs to be checked for two properties:

One-one (Injective): If $f(x_1) = f(x_2)$, then $x_1 = x_2$

Onto (Surjective): Every real number should be achievable as an output

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For $f$ to be onto $\mathbb{R}$, the range must equal $\mathbb{R}$.

Finding $f'(x)$:

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

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

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

For maximum/minimum, set $f'(x) = 0$:

$1-x^2 = 0$

$x = \pm 1$

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At $x = 1$:

$f(1) = \frac{1}{1+1} = \frac{1}{2}$ (maximum)

At $x = -1$:

$f(-1) = \frac{-1}{1+1} = -\frac{1}{2}$ (minimum)

As $x \to \infty$: $f(x) = \frac{x}{x^2+1} \approx \frac{x}{x^2} = \frac{1}{x} \to 0^+$

As $x \to -\infty$: $f(x) \to 0^-$

Therefore, Range $= [-\frac{1}{2}, \frac{1}{2}]$

Since the range is not all of $\mathbb{R}$ (for example, $f(x) = 2$ is never achieved), the function is not onto.

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Testing if $f(x) = \frac{1}{4}$ has multiple solutions:

$\frac{x}{x^2+1} = \frac{1}{4}$

$4x = x^2 + 1$

$x^2 - 4x + 1 = 0$

Using the quadratic formula:

$x = \frac{4 \pm \sqrt{16-4}}{2}$

$x = \frac{4 \pm \sqrt{12}}{2}$

$x = \frac{4 \pm 2\sqrt{3}}{2}$

$x = 2 \pm \sqrt{3}$

Two different $x$-values exist:

$x_1 = 2 + \sqrt{3} \approx 3.73$

$x_2 = 2 - \sqrt{3} \approx 0.27$

Both give $f(x) = \frac{1}{4}$, so different inputs produce the same output.

Therefore, the function is not one-one.

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Since $f(x)$ is neither one-one nor onto, the answer is: neither one-one nor onto.