If `f(x) = sqrt(2-x)` and `g(x) = sqrt(1-2x)`, then the domain of fog (x) is

Options

A. `(-oo, 1//2)`
B. `[1//2, oo)`
C. ` [-3/2,1/2]`
D. none of these

Explanation

To find the domain of the composite function \( f(g(x)) \), we need to follow these steps:

### Step 1: Determine the domain of \( g(x) \)

Given:
\[ g(x) = \sqrt{1 - 2x} \]
For \( g(x) \) to be defined, the expression under the square root must be non-negative:
\[ 1 - 2x \geq 0 \]
Solving this inequality:
\[ 1 \geq 2x \implies x \leq \frac{1}{2} \]
Thus, the domain of \( g(x) \) is:
\[ x \in (-\infty, \frac{1}{2}] \]

### Step 2: Determine the domain of \( f(x) \)

Given:
\[ f(x) = \sqrt{2 - x} \]
For \( f(x) \) to be defined, the expression under the square root must also be non-negative:
\[ 2 - x \geq 0 \]
Solving this inequality:
\[ x \leq 2 \]
Thus, the domain of \( f(x) \) is:
\[ x \in (-\infty, 2] \]

### Step 3: Find the range of \( g(x) \)

Since \( g(x) \) is defined for \( x \leq \frac{1}{2} \), we need to find the maximum value of \( g(x) \) at \( x = \frac{1}{2} \):
\[ g\left(\frac{1}{2}\right) = \sqrt{1 - 2 \cdot \frac{1}{2}} = \sqrt{0} = 0 \]
As \( x \) approaches \(-\infty\), \( g(x) \) approaches \( \sqrt{1} = 1 \). Therefore, the range of \( g(x) \) is:
\[ g(x) \in [0, 1] \]

### Step 4: Determine the domain of \( f(g(x)) \)

Now we need to ensure that \( g(x) \) falls within the domain of \( f(x) \). Since \( f(x) \) is defined for \( x \leq 2 \), we check the range of \( g(x) \):
\[ g(x) \in [0, 1] \implies 0 \leq g(x) \leq 1 \]
Since the entire range of \( g(x) \) is within the domain of \( f(x) \), we can conclude that \( f(g(x)) \) is defined for all \( x \) in the domain of \( g(x) \).

### Step 5: Combine the domains

The domain of \( f(g(x)) \) is the same as the domain of \( g(x) \):
\[ x \in (-\infty, \frac{1}{2}] \]

### Final Answer

Thus, the domain of \( f(g(x)) \) is:
\[ \text{Domain of } f(g(x)) = (-\infty, \frac{1}{2}] \]