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.