Step-by-step Solution:

To find the valid range of \(x\) for which the function is defined, we must find the function's domain. The key principle for this problem is the domain restriction of the inverse cosine function.

1. Understand the Domain of Inverse Cosine:
The function \(y = \cos^{-1}(u)\) (or arccos(u)) is defined only when its argument, \(u\), is within the closed interval \([-1, 1]\).
That is, the condition \(-1 \le u \le 1\) must be met.

2. Apply the Domain Restriction to the Given Function:
For the function \(f(x) = \cos^{-1}(2x-3)\), the argument is \(u = 2x-3\).
Therefore, for the function to be defined, the following inequality must hold true:
\[-1 \le 2x-3 \le 1\]
3. Solve the Inequality for x:
We solve this compound inequality by isolating \(x\) in the middle.
First, add 3 to all three parts of the inequality:
\[-1 + 3 \le 2x - 3 + 3 \le 1 + 3\]\[2 \le 2x \le 4\]
Next, divide all three parts by 2:
\[\frac{2}{2} \le \frac{2x}{2} \le \frac{4}{2}\]\[1 \le x \le 2\]
4. State the Result and Analyze Options:
The valid range for \(x\) is the closed interval \([1, 2]\).

Upon reviewing the given options:
A: \([\frac{1}{3},2]\)
B: \([\frac{1}{3},1]\)
C: [-1,1]
D: \([-\frac{1}{3},\frac{1}{3})\)
The correctly derived domain, \([1, 2]\), does not match any of the provided choices. This indicates that there is a significant error in the question's provided options.

Result:
The correct domain for the function \(\cos^{-1}(2x-3)\) is \([1, 2]\). None of the given options is correct.