Domain of the function
`f(x) = sqrt(3-x) + cos^(-1)((3-2x))/5` is

Options

A. `[-1,3]`
B. `(-1,3)`
C. `[-1,3)`
D. none of these

Answer

To find the domain of the function \( f(x) = \sqrt{3 - x} + \frac{\cos^{-1}(3 - 2x)}{5} \), we need to ensure that both parts of the function are defined.

### Step 1: Analyze the square root function
The expression under the square root, \( 3 - x \), must be non-negative:
\[ 3 - x \geq 0 \]
This simplifies to:
\[ x \leq 3 \]
Thus, we have our first condition:
\[ x \leq 3 \quad \text{(Condition 1)} \]

### Step 2: Analyze the inverse cosine function
The argument of the inverse cosine function, \( 3 - 2x \), must lie within the range of \([-1, 1]\):
\[ -1 \leq 3 - 2x \leq 1 \]

#### Step 2.1: Solve the left inequality
Starting with the left part:
\[ -1 \leq 3 - 2x \]
Subtracting 3 from both sides gives:
\[ -4 \leq -2x \]
Dividing by -2 (and reversing the inequality):
\[ x \leq 2 \quad \text{(Condition 2)} \]

#### Step 2.2: Solve the right inequality
Now, for the right part:
\[ 3 - 2x \leq 1 \]
Subtracting 3 from both sides gives:
\[ -2x \leq -2 \]
Dividing by -2 (and reversing the inequality):
\[ x \geq 1 \quad \text{(Condition 3)} \]

### Step 3: Combine the conditions
Now we have three conditions:
1. \( x \leq 3 \)
2. \( x \leq 2 \)
3. \( x \geq 1 \)

The most restrictive conditions are \( x \leq 2 \) and \( x \geq 1 \). Therefore, the combined domain is:
\[ 1 \leq x \leq 2 \]

### Conclusion
The domain of the function \( f(x) \) is:
\[ \text{Domain: } [1, 2] \]