Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{x^2+1}$ then

Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x}{x^2+1}$.

To check if the function is one-one, find the derivative using the quotient rule:

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

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

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

---

Setting $f'(x) = 0$:

$1-x^2 = 0$

$x = \pm 1$

Analyzing the sign of $f'(x)$:

When $x < -1$: $1-x^2 < 0$, so $f'(x) < 0$ (function is decreasing)

When $-1 < x < 1$: $1-x^2 > 0$, so $f'(x) > 0$ (function is increasing)

When $x > 1$: $1-x^2 < 0$, so $f'(x) < 0$ (function is decreasing)

The function has a minimum at $x = -1$ and a maximum at $x = 1$.

---

Since the function decreases, then increases, then decreases again, some y-values are attained twice.

Check $f(0.5)$ and $f(2)$:

$f(0.5) = \frac{0.5}{0.25+1} = \frac{0.5}{1.25} = 0.4$

$f(2) = \frac{2}{4+1} = \frac{2}{5} = 0.4$

Different inputs give the same output, so $f$ is not one-one.

---

To check if the function is onto, find the range by calculating extreme values:

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)

---

Check the limits:

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

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

---

The function only takes values between $-\frac{1}{2}$ and $\frac{1}{2}$.

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

This is not all of $\mathbb{R}$, so $f$ is not onto.

---

Therefore, $f$ is neither one-one nor onto.