To find the **domain** and **range** of the function \( f(x) = \frac{1}{x^2 + 1} \):

### Domain
The domain consists of all values of \( x \) for which the function is defined. 

In this case, \( x^2 + 1 \) is always positive because \( x^2 \geq 0 \) for all real numbers, and adding 1 makes it strictly greater than zero. This means there are no values of \( x \) that make the denominator zero, so the function is defined for all real numbers.

Thus, the domain is:
\[
\boxed{(-\infty, \infty)}
\]

### Range
The range consists of all possible output values of the function. 

Since \( x^2 \geq 0 \), the smallest value of \( x^2 \) is 0. So, \( x^2 + 1 \geq 1 \). The smallest value that \( f(x) \) can take is:
\[
f(x) = \frac{1}{x^2 + 1} \leq \frac{1}{1} = 1
\]
As \( x^2 \) increases, \( f(x) \) approaches 0 but never reaches 0.

Thus, the range of the function is:
\[
\boxed{(0, 1]}
\]

### Summary:
- **Domain**: \( (-\infty, \infty) \)
- **Range**: \( (0, 1] \)

Would you like more details or have any questions?

### Related questions:
1. How do you find the domain of a rational function in general?
2. What would the graph of \( f(x) = \frac{1}{x^2 + 1} \) look like?
3. How do you determine the range of functions involving square roots or cubes?
4. What happens to the range if the function is \( f(x) = \frac{1}{x^2 - 1} \)?
5. Can you explain the behavior of functions that have horizontal asymptotes?

### Tip:
For rational functions, check for values of \( x \) that make the denominator zero to determine the domain.

Explore how to find the domain and range of the rational function f(x) = 1/(x^2 + 1) with a step-by-step explanation. This problem covers essential algebraic concepts, including rational functions and their behavior, suitable for students in Grades 10-12.

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