To find the range of the function $f(x) = \frac{1}{\sqrt{x + 1}}$, we need to explore the possible values that $f(x)$ can take for all values of $x$ within the domain.

Step 1: Identify the Domain

The expression $\sqrt{x + 1}$ requires that $x + 1 \geq 0$ (since the square root of a negative number is not real). Thus, we must have:

$x + 1 \geq 0 \quad \Rightarrow \quad x \geq -1$

So, the domain of the function is $x \geq -1$.

Step 2: Behavior of the Function

The function can be rewritten as:

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

As $x$ approaches $-1$ from the right (i.e., $x \to -1^+$), the denominator $\sqrt{x + 1}$ approaches 0. Therefore, $f(x) \to \infty$ as $x \to -1^+$.

As $x \to \infty$, $\sqrt{x + 1}$ grows larger, causing $f(x)$ to approach 0. Therefore, $f(x) \to 0^+$ as $x \to \infty$.

Step 3: Range of the Function

Given that $f(x) \to \infty$ as $x \to -1^+$ and $f(x) \to 0^+$ as $x \to \infty$, the function takes all positive real values between 0 and $\infty$. Thus, the range of the function is:

$(0, \infty)$

Final Answer:

The range of $f(x) = \frac{1}{\sqrt{x + 1}}$ is $(0, \infty)$.

Would you like a more detailed explanation, or do you have any questions about this problem? Here are some related questions to explore:

1. How would the range change if the function were $f(x) = \frac{1}{\sqrt{x - 1}}$?
2. What happens if we modify the function to $f(x) = \frac{1}{\sqrt{x + 1}} + 2$?
3. How do transformations like translations and scalings affect the range of a function?
4. Can you explain how the domain of a function affects its range?
5. What would the range be if we had the function $f(x) = \frac{1}{\sqrt{x + 1}}$ for $x \in (-1, 3)$?

Tip: When analyzing the range of a function, start by identifying its domain and understanding the behavior of the function as it approaches boundary points (like limits).