It seems like you're working with a function, $f(x) = -2 \sqrt{3 - x} + 5$. Let's break down the components and analyze the function.

This function involves a square root, which indicates some potential constraints for the domain. Specifically, the expression inside the square root, $3 - x$, must be greater than or equal to zero, because the square root of a negative number is undefined in the real number system.

Step 1: Determine the domain

For the square root $\sqrt{3 - x}$ to be valid, the inside of the square root, $3 - x$, must satisfy: $3 - x \geq 0$ Solving this inequality: $x \leq 3$ Thus, the domain of $f(x)$ is all $x$ such that $x \leq 3$, or in interval notation, $(-\infty, 3]$.

Step 2: Behavior of the function

The function involves a negative coefficient for the square root, so the value of $f(x)$ will decrease as $x$ increases. The $+5$ at the end translates the function vertically by 5 units.

Step 3: Finding key values

To explore the behavior further, we can plug in specific values of $x$. Let's check:

• When $x = 3$, we get: $f(3) = -2 \sqrt{3 - 3} + 5 = -2(0) + 5 = 5$

• When $x = 0$, we get: $f(0) = -2 \sqrt{3 - 0} + 5 = -2 \sqrt{3} + 5$

Would you like to go over further evaluations, graphing, or other details related to this function?

---

Here are 5 related questions to consider:

1. How do we graph this function?
2. What happens to the value of $f(x)$ as $x \to -\infty$?
3. Can this function be differentiated? If so, what's the derivative?
4. How does the behavior change if the coefficient of the square root were positive?
5. What are the limits of $f(x)$ as $x$ approaches 3 from the left?

Tip: Always consider the domain when working with functions involving square roots, as they place constraints on the input values.