To find the domain and range of the given function:

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

Domain:

The domain of a function is the set of all possible input values (x) that make the function well-defined. In this case, the function has a denominator, and we must ensure that the denominator is not zero.

The denominator $x + 2$ is zero when: $x + 2 = 0 \implies x = -2$ Thus, the function is undefined when $x = -2$.

Therefore, the domain is all real numbers except $x = -2$, which can be written as: $\text{Domain}: \{ x \in \mathbb{R} \mid x \neq -2 \}$ or in interval notation: $\text{Domain}: (-\infty, -2) \cup (-2, +\infty)$

Range:

The range of a function is the set of all possible output values (f(x)) for the domain.

For the given function $f(x) = \frac{1}{x + 2}$, the output will never be zero because no value of $x$ makes $\frac{1}{x + 2} = 0$.

Also, as $x$ approaches $-2$, the value of $f(x)$ approaches infinity or negative infinity, depending on the direction from which $x$ approaches.

Therefore, the range includes all real numbers except $f(x) = 0$, which can be written as: $\text{Range}: \{ f(x) \in \mathbb{R} \mid f(x) \neq 0 \}$ or in interval notation: $\text{Range}: (-\infty, 0) \cup (0, +\infty)$

Conclusion:

• Domain: $(-\infty, -2) \cup (-2, +\infty)$
• Range: $(-\infty, 0) \cup (0, +\infty)$

Would you like further details or clarifications?

Here are some related questions to expand on this topic:

1. How would the domain and range change for $f(x) = \frac{1}{x - 3}$?
2. What happens to the domain and range if the function is $f(x) = \frac{2}{x + 2}$?
3. Can a function have an empty domain or range?
4. How does the graph of $f(x) = \frac{1}{x+2}$ look?
5. What is the domain and range of $f(x) = \sqrt{x + 2}$?

Tip: Always check for restrictions on the domain, such as division by zero or square roots of negative numbers, to avoid undefined points.