The function given is:

$f(x) = \left|\sqrt{x^2 - 1}\right| + 4$

To find the range of this function, let's analyze it step by step.

Step 1: Analyze the expression inside the absolute value

The expression inside the square root is $x^2 - 1$. Since the square root function is defined only for non-negative values, we have:

$x^2 - 1 \geq 0$

This inequality holds when:

$x^2 \geq 1 \quad \Rightarrow \quad |x| \geq 1 \quad \Rightarrow \quad x \leq -1 \text{ or } x \geq 1$

So, the function $f(x)$ is defined for $x \leq -1$ and $x \geq 1$.

Step 2: Simplify the function

Since $x^2 - 1 \geq 0$, the square root is always non-negative. Thus, the absolute value $\left|\sqrt{x^2 - 1}\right|$ simply equals $\sqrt{x^2 - 1}$. Therefore:

$f(x) = \sqrt{x^2 - 1} + 4$

Step 3: Determine the minimum and maximum values of $f(x)$

The minimum value of $\sqrt{x^2 - 1}$ occurs when $x^2$ is at its minimum within the domain $|x| \geq 1$. For $|x| = 1$, we have:

$\sqrt{1^2 - 1} = \sqrt{0} = 0$

So, the minimum value of $f(x)$ is:

$f(x)_{\text{min}} = 0 + 4 = 4$

As $|x|$ increases (i.e., $|x| > 1$), $x^2$ increases, and thus $\sqrt{x^2 - 1}$ increases as well. Since the square root function tends to infinity as $x^2$ becomes large, $f(x)$ has no upper bound.

Conclusion

The range of $f(x)$ is:

$\text{Range of } f(x) = [4, \infty)$

Would you like more details or have any questions? Here are some related questions you might consider:

1. How would the range of $f(x)$ change if we subtracted 4 instead of adding it?
2. What is the domain of the function $f(x)$?
3. How would the range be affected if the function had $\sqrt{1 - x^2}$ instead?
4. What are the properties of square root functions and their impact on range?
5. How would the range be different if the absolute value were applied to the entire expression?

Tip: When finding the range of a function, consider the behavior of the function at the boundaries of its domain as well as the function's tendency (e.g., does it go to infinity or approach a certain value?).