Find the domain of following function: `f(x)=sqrt(1-x)-"sin^(-1)((2x-1)/3)`

Explanation

To find the domain of the function \( f(x) = \sqrt{1-x} - \sin^{-1}\left(\frac{2x-1}{3}\right) \), we need to determine the restrictions imposed by both components of the function: the square root and the inverse sine.

### Step 1: Determine the domain of \( \sqrt{1-x} \)

The expression inside the square root must be non-negative:
\[ 1 - x \geq 0 \]
This simplifies to:
\[ x \leq 1 \]
Thus, the domain for the square root function is:
\[ (-\infty, 1] \]

### Step 2: Determine the domain of \( \sin^{-1}\left(\frac{2x-1}{3}\right) \)

The argument of the inverse sine function must lie within the interval \([-1, 1]\):
\[ -1 \leq \frac{2x-1}{3} \leq 1 \]

#### Sub-step 2.1: Solve the left inequality
\[ -1 \leq \frac{2x-1}{3} \]
Multiplying both sides by 3:
\[ -3 \leq 2x - 1 \]
Adding 1 to both sides:
\[ -2 \leq 2x \]
Dividing by 2:
\[ -1 \leq x \]

#### Sub-step 2.2: Solve the right inequality
\[ \frac{2x-1}{3} \leq 1 \]
Multiplying both sides by 3:
\[ 2x - 1 \leq 3 \]
Adding 1 to both sides:
\[ 2x \leq 4 \]
Dividing by 2:
\[ x \leq 2 \]

### Step 3: Combine the results

From the square root, we found \( x \leq 1 \), and from the inverse sine, we found:
\[ -1 \leq x \leq 2 \]

Now, we need to find the intersection of these two domains:
- From the square root: \( (-\infty, 1] \)
- From the inverse sine: \( [-1, 2] \)

The intersection is:
\[ [-1, 1] \]

### Conclusion

Thus, the domain of the function \( f(x) \) is:
\[ \boxed{[-1, 1]} \]