To determine the behavior of the function $h(x) = 3\sqrt{x}$, we need to analyze its derivative. The derivative of $h(x)$ will tell us where the function is increasing or decreasing.

First, let's rewrite the function $h(x)$ in a more convenient form for differentiation: $h(x) = 3\sqrt{x} = 3x^{1/2}$

Next, we find the derivative $h'(x)$: $h'(x) = 3 \cdot \frac{1}{2} x^{-1/2} = \frac{3}{2} x^{-1/2} = \frac{3}{2\sqrt{x}}$

Now, let's analyze the sign of $h'(x)$ to determine where the function is increasing or decreasing. The key is to note that the derivative $h'(x)$ is defined and positive only for $x > 0$. For $x < 0$, the function $3\sqrt{x}$ is not real-valued because the square root of a negative number is not defined in the set of real numbers.

• For $x > 0$: The derivative $h'(x) = \frac{3}{2\sqrt{x}}$ is positive. Since the derivative is positive, the function $h(x)$ is increasing on the interval $(0, \infty)$.

• For $x \leq 0$: The function $h(x)$ is not defined in the real numbers.

Given this information, the correct statement is:

• The function is increasing on the interval $(0, \infty)$.

Would you like more details or have any questions?

Here are 5 additional questions you might ask:

1. How do you find the derivative of a function?
2. What does it mean for a function to be increasing or decreasing?
3. Can you explain the concept of intervals in calculus?
4. What are the common rules for differentiating functions?
5. How do you determine if a function is defined on a certain interval?

Tip: When analyzing a function's behavior, always check the domain to ensure the function is defined and real-valued in the interval of interest.