find the domain and the range of the function below: g(x) = x+3/x^2+1 

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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find the domain and the range of the function below: g(x) = (x+3)/(x^2+1)

Explore how to find the domain and range of the rational function g(x) = (x + 3)/(x^2 + 1). This problem demonstrates key concepts in algebra, including limits and behavior at infinity, ideal for students in Grades 10-12.

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