We are given the function:

$g(x) = \frac{x + 3}{x^2 + 1}$

1. Finding the Domain

The domain of a function consists of all possible $x$-values for which the function is defined. For rational functions like $g(x)$, the main issue is division by zero. So, we check if the denominator can ever be zero.

The denominator is $x^2 + 1$. This expression is never zero because:

$x^2 + 1 \geq 1 \quad \text{for all real values of } x$

Thus, the denominator is always positive and non-zero. Therefore, there are no restrictions on $x$, meaning that the domain is:

$\text{Domain of } g(x): \quad (-\infty, \infty)$

2. Finding the Range

To find the range, we need to determine the possible values of $g(x)$. The function:

$g(x) = \frac{x + 3}{x^2 + 1}$

We can analyze the behavior of the function as $x \to \infty$ and $x \to -\infty$:

• As $x \to \infty$, the $x^2$ term dominates in the denominator, so:

$g(x) \approx \frac{x}{x^2} = \frac{1}{x} \quad \Rightarrow \quad g(x) \to 0$

• As $x \to -\infty$, the same logic applies, so $g(x) \to 0$ as $x \to -\infty$.

Thus, the function approaches $0$, but never reaches it.

Next, we check specific values to see if other values are possible. For example:

• At $x = 0$:

$g(0) = \frac{0 + 3}{0^2 + 1} = 3$

• For large positive and negative values of $x$, $g(x)$ can get arbitrarily close to $0$, but never reach it.

Therefore, the function covers values from $0$ (but not including $0$) up to a maximum of 3 (or even higher depending on exact values). Based on this, the range is:

$\text{Range of } g(x): \quad (-\infty, \infty)$

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