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 argument of the inverse sine function lies within its valid range. The domain of \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, we need to solve the inequality:

\[ -1 \leq 4x + 3 \leq 1 \]

### Step 1: Break the inequality into two parts
We can break this compound inequality into two separate inequalities:

1. \( 4x + 3 \geq -1 \)
2. \( 4x + 3 \leq 1 \)

### Step 2: Solve the first inequality
Starting with the first inequality:

\[ 4x + 3 \geq -1 \]

Subtract 3 from both sides:

\[ 4x \geq -1 - 3 \]
\[ 4x \geq -4 \]

Now, divide both sides by 4:

\[ x \geq -1 \]

### Step 3: Solve the second inequality
Now, let's solve the second inequality:

\[ 4x + 3 \leq 1 \]

Subtract 3 from both sides:

\[ 4x \leq 1 - 3 \]
\[ 4x \leq -2 \]

Now, divide both sides by 4:

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

### Step 4: Combine the results
Now we combine the results from both inequalities:

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

### Conclusion
Thus, the domain of the function \( f(x) = \sin^{-1}(4x + 3) \) is:

\[ [-1, -\frac{1}{2}] \]