What is the domain of f(x)=`3sqrt(15-x^2)`?

Options

A. xgt0
B. xgt2.47
C. -2.47ltxlt2.47
D. all real numbers

Explanation

To find the domain of the function \( f(x) = \sqrt[3]{15 - x^2} \), we need to determine the values of \( x \) for which the function is defined.

### Step-by-Step Solution:

1. **Identify the Function**: The function is given as \( f(x) = \sqrt[3]{15 - x^2} \). This indicates that we are dealing with a cube root function.

2. **Understanding Cube Roots**: The cube root function, \( \sqrt[3]{y} \), is defined for all real numbers \( y \). This means that there are no restrictions on the value of \( y \) for the cube root to be real.

3. **Set Up the Inequality**: We need to ensure that the expression inside the cube root, \( 15 - x^2 \), can take any real value. There are no restrictions on \( 15 - x^2 \) since it can be negative, zero, or positive.

4. **Determine the Range of \( x \)**: Since \( 15 - x^2 \) can take any real number value, we can conclude that \( x^2 \) can take values from \( -\infty \) to \( 15 \). However, since \( x^2 \) is always non-negative, we can say that \( x^2 \) can range from \( 0 \) to \( 15 \).

5. **Conclusion on Domain**: Since \( x^2 \) can take any value from \( 0 \) to \( 15 \), and since \( x \) can be any real number, the domain of \( f(x) \) is all real numbers. Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = (-\infty, \infty) \]

### Final Answer:
The domain of \( f(x) = \sqrt[3]{15 - x^2} \) is all real numbers, \( (-\infty, \infty) \).