The domain of definition of the function
`f(x)=sin^(-1)((x-3)/(2))-log_(10)(4-x)` , is

Options

A. `1 le x le 5`
B. `1 lt x lt 4`
C. `1 le x lt 4`
D. `1 le x le 4`

Answer

To find the domain of the function \( f(x) = \sin^{-1}\left(\frac{x-3}{2}\right) - \log_{10}(4-x) \), we need to ensure that both parts of the function are defined.

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

The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to solve the inequality:

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

#### Solving the left part of the inequality:

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

Multiplying both sides by 2:

\[ -2 \leq x - 3 \]

Adding 3 to both sides:

\[ 1 \leq x \quad \text{or} \quad x \geq 1 \]

#### Solving the right part of the inequality:

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

Multiplying both sides by 2:

\[ x - 3 \leq 2 \]

Adding 3 to both sides:

\[ x \leq 5 \]

### Step 2: Combine the results from the first part

From the first part, we have:

\[ 1 \leq x \leq 5 \]

### Step 3: Determine the domain for \( -\log_{10}(4-x) \)

The logarithmic function \( \log_{10}(y) \) is defined for \( y > 0 \). Therefore, we need:

\[ 4 - x > 0 \]

This simplifies to:

\[ x < 4 \]

### Step 4: Combine the results from both parts

Now we combine the two conditions:

1. From \( \sin^{-1}\left(\frac{x-3}{2}\right) \): \( 1 \leq x \leq 5 \)
2. From \( -\log_{10}(4-x) \): \( x < 4 \)

The intersection of these two conditions is:

\[ 1 \leq x < 4 \]

### Final Domain

Thus, the domain of the function \( f(x) \) is:

\[ [1, 4) \]