We are tasked with determining the intervals where the function $f(x) = \sqrt{x^2 - 9}$ is increasing, decreasing, or constant. Let's analyze this step by step.

Step 1: Domain of the Function

The square root function is defined when the argument is non-negative: $x^2 - 9 \geq 0 \implies x \leq -3 \text{ or } x \geq 3.$ Thus, the domain of $f(x)$ is $(-\infty, -3] \cup [3, \infty)$.

Step 2: Analyze the Behavior

The graph provided shows:

1. The function decreases as $x$ moves from $-\infty$ to $-3$.
2. The function increases as $x$ moves from $3$ to $\infty$.

Between $-3$ and $3$, the function is not defined, as $x^2 - 9 < 0$ in this interval.

Step 3: Final Answer

• Increasing: $(3, \infty)$
• Decreasing: $(-\infty, -3)$
• Constant: DNE (Does Not Exist)

Would you like a detailed explanation of the derivative analysis or have any further questions?

Here are 5 related questions to expand your understanding:

1. How do we compute the derivative of $f(x) = \sqrt{x^2 - 9}$?
2. What happens to the slope of the tangent line as $x$ approaches $-3$ or $3$?
3. Can we determine concavity for this function? If so, how?
4. How do domain restrictions affect the increasing/decreasing behavior of functions?
5. What is the significance of critical points in identifying increasing and decreasing intervals?

Tip: Always confirm the domain of a function before analyzing its behavior; it determines where the function is valid and impacts increasing/decreasing intervals.