Final answer:
The function f(x) = 3(\sqrt{18})^x has a domain of all real numbers, an initial value of 3, and a simplified base of 3\sqrt{2}. The range is y > 0, and the statement about the initial value being 9 is incorrect.
Step-by-step explanation:
The function in question is f(x) = 3(\sqrt{18})^x. To correctly describe this function, we need to focus on details such as domain, range, initial value, and the nature of the base of the exponential term.
- The domain of any exponential function like this one is all real numbers, because we can plug any real number in for x and the function will yield a real result.
- The range of this function is y > 0, because the output of the function will always be positive. This is a property of exponential functions with positive bases. Thus, the statement "The range is y > 3" is incorrect.
- The initial value of an exponential function is the value when x is zero. So, for this function, when x = 0, f(x) = 3\sqrt{18}^0 = 3, meaning the initial value is indeed 3.
- Simplifying the base, we have \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}, so the simplified base is 3\sqrt{2}.
- The statement that "The initial value is 9" is incorrect, as already established that the initial value is 3.
Summing up, the correct descriptions are: the domain is all real numbers, the initial value is 3, and the simplified base is 3\sqrt{2}.