If `f(x)=sin^(-1)(4x+3)`, then the domain of f is:

Options

A. `(-infty,0]`
B. `[-1,(-1)/2]`
C. `[-1,infty)`
D. `(-infty, infty)`

Answer

To find the domain of the function \( f(x) = \sin^{-1}(4x + 3) \), we need to ensure that the expression inside the inverse sine function, \( 4x + 3 \), lies within the valid range for the inverse sine function. The domain of \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\).

### Step-by-step Solution:

1. **Set up the inequality**: We need to find the values of \( x \) such that:
\[ -1 \leq 4x + 3 \leq 1 \]

2. **Solve the left part of the inequality**:
\[ -1 \leq 4x + 3 \]
Subtract 3 from both sides:
\[ -1 - 3 \leq 4x \]
Simplifying gives:
\[ -4 \leq 4x \]
Now, divide both sides by 4:
\[ -1 \leq x \]

3. **Solve the right part of the inequality**:
\[ 4x + 3 \leq 1 \]
Subtract 3 from both sides:
\[ 4x \leq 1 - 3 \]
Simplifying gives:
\[ 4x \leq -2 \]
Now, divide both sides by 4:
\[ x \leq -\frac{1}{2} \]

4. **Combine the results**: From the two parts, we have:
\[ -1 \leq x \leq -\frac{1}{2} \]
This can be written in interval notation as:
\[ x \in [-1, -\frac{1}{2}] \]

### Final Answer:
The domain of \( f(x) = \sin^{-1}(4x + 3) \) is:
\[ [-1, -\frac{1}{2}] \]