The function you provided is $f(x) = \frac{2x + x}{x}$. We can simplify this function before determining the domain and range.

Step 1: Simplify the function

$f(x) = \frac{2x + x}{x} = \frac{3x}{x} = 3 \quad \text{(for } x \neq 0\text{)}.$ Thus, $f(x) = 3$ for all $x \neq 0$.

Step 2: Domain

The domain of a function is the set of all possible inputs (values of $x$) for which the function is defined. Since the function $f(x) = \frac{3x}{x}$ has $x$ in the denominator, the only value that makes the denominator zero is $x = 0$. Therefore, $f(x)$ is undefined at $x = 0$.

Domain: $x \in (-\infty, 0) \cup (0, \infty)$.

Step 3: Range

The range of a function is the set of all possible outputs (values of $f(x)$). Since the function simplifies to $f(x) = 3$ for all $x \neq 0$, the output is always 3, except for $x = 0$, where the function is undefined.

Range: $\{ 3 \}$.

Final Answer:

• Domain: $(-\infty, 0) \cup (0, \infty)$
• Range: $\{ 3 \}$

Would you like further details or have any questions?

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Here are 5 related questions for practice:

1. What is the domain and range of $f(x) = \frac{1}{x}$?
2. How does the domain of $f(x) = \frac{x^2 + x}{x}$ differ from your original function?
3. If $f(x) = \frac{2x+1}{x-1}$, what is its domain and range?
4. What happens if you multiply both numerator and denominator by $x^2$?
5. How does removing the denominator affect the domain of a rational function?

Tip: Always check for values that make the denominator zero to find restrictions on the domain of rational functions.