Step 1: Understanding the Concept:
To determine the range of the composite function \( f(x) \), we examine how the inner expression \( \frac{\pi}{\sqrt{x+1}+4} \) varies within the allowed values of \( x \), and then observe how those values behave when passed through the tangent function.
Step 2: Key Formula or Approach:
1. For a square root expression \( \sqrt{g(x)} \), the condition \( g(x) \geq 0 \) must hold.
2. The range of the function \( \sqrt{x} \) is \( [0, \infty) \).
3. The tangent function \( \tan \theta \) increases strictly over the interval \( (0, \frac{\pi}{2}) \).
Step 3: Detailed Explanation:
First, identify the domain of \( f(x) \). For the square root to exist:
\[ x+1 \geq 0 \implies x \geq -1 \]
Next, consider the expression inside the tangent function. Let
\[ \theta = \frac{\pi}{\sqrt{x+1}+4} \]
Since \( \sqrt{x+1} \ge 0 \) for all \( x \ge -1 \), we have:
\[ \sqrt{x+1} + 4 \ge 4 \]
Taking the reciprocal reverses the inequality because all quantities are positive:
\[ 0 \textless \frac{1}{\sqrt{x+1}+4} \le \frac{1}{4} \]
Multiplying through by \( \pi \):
\[ 0 \textless \frac{\pi}{\sqrt{x+1}+4} \le \frac{\pi}{4} \]
Therefore, the angle \( \theta \) belongs to the interval \( (0, \frac{\pi}{4}] \).
Now evaluate \( f(x) = \tan \theta \) for \( \theta \in (0, \frac{\pi}{4}] \).
Because \( \tan \theta \) is increasing on this interval:
\[ \lim_{\theta \to 0^+} \tan \theta \textless f(x) \le \tan \left( \frac{\pi}{4} \right) \]
\[ 0 \textless f(x) \le 1 \]
Hence, the range of \( f \) is \( (0, 1] \).
Step 4: Final Answer:
The range of the function is \( (0, 1] \).