The function $f: [-1, 1] \rightarrow R$ (set of real numbers) given by $f(x) = \frac{x}{x+3}$ is

The function $f: [-1, 1] \to \mathbb{R}$ is given by $f(x) = \frac{x}{x+3}$.

To determine if the function is one-one (injective) and onto (surjective):

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

$f'(x) = \frac{d}{dx}\left(\frac{x}{x+3}\right)$

Using the quotient rule:

$f'(x) = \frac{(x+3)(1) - x(1)}{(x+3)^2}$

$f'(x) = \frac{x+3-x}{(x+3)^2}$

$f'(x) = \frac{3}{(x+3)^2}$

For $x \in [-1, 1]$, we have $x+3 \in [2, 4]$, so $(x+3)^2 > 0$.

Therefore $f'(x) = \frac{3}{(x+3)^2} > 0$ for all $x \in [-1, 1]$.

The function is strictly increasing on $[-1, 1]$, so it is one-one.

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To check if the function is onto, find the range.

Since $f$ is continuous and increasing on $[-1, 1]$, the range is $[f(-1), f(1)]$.

$f(-1) = \frac{-1}{-1+3}$

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

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

$f(1) = \frac{1}{1+3}$

$f(1) = \frac{1}{4}$

The range of $f$ is $\left[-\frac{1}{2}, \frac{1}{4}\right]$.

The codomain is $\mathbb{R}$ (all real numbers).

Since the range $\left[-\frac{1}{2}, \frac{1}{4}\right] \neq \mathbb{R}$, the function is not onto.

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The function is one-one but not onto.