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To find the **domain** of the function  
\[
f(x) = \frac{\sqrt{x+1}}{x^2 + 2x},
\]
we analyze where the function is **defined**:

1. **Square Root Condition**: The expression under the square root must be **non-negative**:
   \[
   x+1 \geq 0 \implies x \geq -1.
   \]

2. **Denominator Condition**: The denominator cannot be **zero** because division by zero is undefined:
   \[
   x^2 + 2x \neq 0.
   \]
   Factorizing:
   \[
   x(x+2) \neq 0 \implies x \neq 0 \quad \text{and} \quad x \neq -2.
   \]

---

### Combining the Conditions:
1. From the square root condition: \( x \geq -1 \).
2. From the denominator: \( x \neq 0 \) and \( x \neq -2 \).

Since \( x \geq -1 \) excludes values less than \(-1\), we **ignore** \( x = -2 \) because it is already outside the range \( x \geq -1 \).

Thus, the restrictions become:
\[
x \geq -1 \quad \text{and} \quad x \neq 0.
\]

---

### Final Domain:
The domain of \( f(x) \) is:
\[
x \in [-1, 0) \cup (0, \infty).
\]

This means \( x \) can take values starting from \(-1\) to \( 0 \) (excluding \( 0 \)) and then all values greater than \( 0 \).

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### Would you like further details or a graphical explanation? Let me know!  

Here are 5 related questions for practice:
1. Find the domain of \( g(x) = \frac{\sqrt{2x - 3}}{x^2 - 5x} \).
2. Determine the range of \( f(x) = \frac{\sqrt{x+1}}{x^2 + 2x} \).
3. Solve \( x^2 + 2x = 0 \) for its roots.
4. What are the conditions for the square root \( \sqrt{f(x)} \) to be defined?
5. Analyze the asymptotes of \( h(x) = \frac{1}{x^2 + 2x} \).

**Tip**: Always analyze functions involving square roots and denominators for restrictions carefully!

Find the domain of f(x) = (√(x+1)) / (x^2 + 2x).

Learn how to determine the domain of the function f(x) = (√(x+1)) / (x² + 2x). This solution analyzes square root restrictions and rational function denominators, step-by-step, suitable for Grades 9-12.

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