To determine whether the function \( g(x) = \sqrt{x}^3 \) is even, odd, or neither, we need to check the definitions of even and odd functions:
- Even function: A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \).
- Odd function: A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \).
Now, let's analyze the function \( g(x) = \sqrt{x}^3 \).
First, we simplify the expression:
\[ g(x) = (\sqrt{x})^3 = x^{3/2}. \]
Next, we find \( g(-x) \):
\[ g(-x) = \sqrt{-x}^3. \]
However, \( \sqrt{-x} \) is not defined for real numbers when \( x \) is positive (making the domain restrictive). Thus, we can only evaluate \( g(-x) \) for \( x < 0 \).
For negative \( x \):
- Since \( g(-x) \) involves taking the square root of a negative number, \( g(-x) \) is not defined in the set of real numbers.
- Since \( g(x) \) is not defined for negative \( x \), we cannot demonstrate either the evenness or oddness of \( g(x) \).
Thus, we conclude that \( g(x) \) is neither even nor odd.
Final answer: neither even nor odd.