`f(x)=sin^(-1)((3-2x)/5)+sqrt(3-x)`.Find the domain of f(x).

Answer

To find the domain of the function \( f(x) = \sin^{-1}\left(\frac{3-2x}{5}\right) + \sqrt{3-x} \), we need to determine the individual domains of the two components of the function, \( f_1(x) = \sin^{-1}\left(\frac{3-2x}{5}\right) \) and \( f_2(x) = \sqrt{3-x} \), and then find their intersection.

### Step 1: Determine the domain of \( f_1(x) \)

The function \( \sin^{-1}(x) \) is defined for \( -1 \leq x \leq 1 \). Therefore, we need to set up the inequalities for \( \frac{3-2x}{5} \):

\[ -1 \leq \frac{3-2x}{5} \leq 1 \]

#### Solving the left inequality:

1. Multiply all parts by 5:
\[ -5 \leq 3 - 2x \]
2. Subtract 3 from all sides:
\[ -8 \leq -2x \]
3. Divide by -2 (remember to reverse the inequality):
\[ 4 \geq x \quad \text{or} \quad x \leq 4 \]

#### Solving the right inequality:

1. Multiply all parts by 5:
\[ 3 - 2x \leq 5 \]
2. Subtract 3 from all sides:
\[ -2x \leq 2 \]
3. Divide by -2 (reverse the inequality):
\[ x \geq -1 \]

Combining both results, we find:
\[ -1 \leq x \leq 4 \]

### Step 2: Determine the domain of \( f_2(x) \)

The function \( \sqrt{3-x} \) is defined when the expression under the square root is non-negative:
\[ 3 - x \geq 0 \]
This simplifies to:
\[ x \leq 3 \]

### Step 3: Find the intersection of the domains

Now we have:
- Domain of \( f_1(x) \): \( -1 \leq x \leq 4 \)
- Domain of \( f_2(x) \): \( x \leq 3 \)

The intersection of these two domains is:
\[ -1 \leq x \leq 3 \]

### Final Answer

Thus, the domain of \( f(x) \) is:
\[ \boxed{[-1, 3]} \]