The function $f: [-1, 1] \rightarrow R$ is given by $f(x) = \frac{x}{x + 2}$

The function $f: [-1, 1] \rightarrow \mathbb{R}$ is given by $f(x) = \frac{x}{x + 2}$

Domain: $[-1, 1]$

Codomain: $\mathbb{R}$

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To check if the function is one-one, assume $f(x_1) = f(x_2)$

$\frac{x_1}{x_1 + 2} = \frac{x_2}{x_2 + 2}$

$x_1(x_2 + 2) = x_2(x_1 + 2)$

$x_1 x_2 + 2x_1 = x_2 x_1 + 2x_2$

$2x_1 = 2x_2$

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function is one-one.

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To check if the function is onto, find the range of $f(x)$ for $x \in [-1, 1]$.

At $x = -1$:

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

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

$f(-1) = -1$

At $x = 1$:

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

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

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To determine if $f(x)$ is monotonic, find the derivative:

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

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

Since $(x+2)^2 > 0$ for all $x \in [-1, 1]$, we have $f'(x) > 0$.

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

Therefore, the range of $f$ is $\left[-1, \frac{1}{3}\right]$.

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Range: $\left[-1, \frac{1}{3}\right]$

Codomain: $\mathbb{R}$

Since the range does not equal the codomain, the function is not onto.

For example, there is no $x \in [-1, 1]$ such that $f(x) = 2$.

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The function is one-one only.