To find the domain of the function $g(x) = \frac{3x}{x^2 + 9}$, we need to determine the values of $x$ for which the function is defined. The function will be undefined wherever the denominator is equal to zero.

Step 1: Identify the denominator

The denominator of the function is $x^2 + 9$.

Step 2: Determine where the denominator equals zero

We solve the equation:

$x^2 + 9 = 0$

Subtract 9 from both sides:

$x^2 = -9$

Step 3: Analyze the equation

Notice that $x^2 = -9$ has no real solutions because the square of any real number cannot be negative. Therefore, the denominator $x^2 + 9$ is never zero for any real number $x$.

Step 4: Determine the domain

Since the denominator is never zero, the function $g(x) = \frac{3x}{x^2 + 9}$ is defined for all real numbers $x$.

Final Answer:

The domain of $g(x)$ is:

$(-\infty, \infty)$

Would you like more details or have any questions?

Related Questions:

1. What is the range of the function $g(x) = \frac{3x}{x^2 + 9}$?
2. Can you find the horizontal asymptote of this function?
3. How would you determine the x-intercepts of the function?
4. What happens to $g(x)$ as $x$ approaches infinity?
5. Can you graph the function to confirm the domain and analyze its behavior?

Tip:

When analyzing the domain of a rational function, always start by checking where the denominator is zero, as these values are not in the domain.