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

Explanation

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

### Step 1: Finding the domain of \( \sqrt{1 - 2x} \)

The expression under the square root must be non-negative:
\[ 1 - 2x \geq 0 \]
To solve this inequality, we can rearrange it:
\[ 1 \geq 2x \]
Dividing both sides by 2 gives:
\[ \frac{1}{2} \geq x \quad \text{or} \quad x \leq \frac{1}{2} \]

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

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

#### Solving the left inequality:
\[ -1 \leq \frac{3x - 1}{2} \]
Multiplying both sides by 2:
\[ -2 \leq 3x - 1 \]
Adding 1 to both sides:
\[ -1 \leq 3x \]
Dividing by 3:
\[ -\frac{1}{3} \leq x \quad \text{or} \quad x \geq -\frac{1}{3} \]

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

### Step 3: Combining the results

From the first component, we found:
\[ x \leq \frac{1}{2} \]

From the second component, we found:
\[ -\frac{1}{3} \leq x \leq 1 \]

Now, we need to find the intersection of these two results:
1. The first condition gives us \( x \leq \frac{1}{2} \).
2. The second condition gives us \( -\frac{1}{3} \leq x \leq 1 \).

The intersection of these two intervals is:
\[ -\frac{1}{3} \leq x \leq \frac{1}{2} \]

### Final Domain

Thus, the domain of the function \( f(x) \) is:
\[ \boxed{\left[-\frac{1}{3}, \frac{1}{2}\right]} \]