https://file.aiquickdraw.com/mathbot/file/17254125043528ygh2d4y.jpg

The expression shown in the image is:

\[
f.H(x) = \frac{\sqrt{x^2 - 4}}{x^2 + 1}
\]

### Steps to Analyze the Function:

1. **Domain**: 
   - The expression inside the square root, \(x^2 - 4\), must be non-negative. Therefore, \(x^2 - 4 \geq 0\) implies \(x^2 \geq 4\). This gives \(x \leq -2\) or \(x \geq 2\).
   - The denominator \(x^2 + 1\) is always positive and non-zero for all real values of \(x\), so it doesn't impose any further restrictions.
   - Hence, the domain of \(f.H(x)\) is \(x \in (-\infty, -2] \cup [2, \infty)\).

2. **Behavior at Critical Points**:
   - Analyze the function at critical points \(x = \pm 2\) and check the limits as \(x\) approaches these points.

3. **Asymptotic Behavior**:
   - Consider the limits as \(x \rightarrow \infty\) and \(x \rightarrow -\infty\).

4. **Simplifications**:
   - Investigate if there are any algebraic simplifications or possible factorizations.

Would you like to proceed with a detailed analysis of these steps? Additionally, here are some questions to explore further:

1. What is the range of the function \(f.H(x)\)?
2. How does \(f.H(x)\) behave as \(x\) approaches \(\pm \infty\)?
3. What are the critical points of \(f.H(x)\)?
4. How does \(f.H(x)\) change when \(x = 0\)?
5. What is the behavior of \(f.H(x)\) near the boundaries of its domain?

**Tip:** Always start by determining the domain of a function before attempting to solve or simplify it. This helps in understanding the behavior of the function and avoiding errors in interpretation.

Analyze the function f.H(x) = (sqrt(x^2 - 4))/(x^2 + 1) with respect to its domain, critical points, and behavior at infinity.

Explore the detailed analysis of the function f.H(x) = sqrt(x^2 - 4)/(x^2 + 1), including its domain, critical points, and behavior at infinity. This solution is ideal for undergraduate mathematics students.

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