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 $0$.

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!