The domain of the function `f (x) =sqrt(x- sqrt(1-x^2))` is

Options

A. `[ -1,-(1)/(sqrt(2))]uu [(1)/(sqrt(2)),1]`
B. `[(1)/(sqrt(2),1)]`
C. `(-oo , (1)/(2) ] uu [1/2 ,oo)`
D. `[-1,1]`

Answer

To find the domain of the function \( f(x) = \sqrt{x - \sqrt{1 - x^2}} \), we need to ensure that the expressions inside the square roots are non-negative.

### Step 1: Analyze the inner square root
The first condition we need to satisfy is:
\[ 1 - x^2 \geq 0 \]
This implies:
\[ x^2 \leq 1 \]
Taking the square root of both sides gives:
\[ -1 \leq x \leq 1 \]

### Step 2: Analyze the outer square root
Next, we need to ensure that the expression \( x - \sqrt{1 - x^2} \) is also non-negative:
\[ x - \sqrt{1 - x^2} \geq 0 \]
This can be rearranged to:
\[ x \geq \sqrt{1 - x^2} \]
Now, squaring both sides (noting that both sides are non-negative within the domain we found) gives:
\[ x^2 \geq 1 - x^2 \]
Combining like terms results in:
\[ 2x^2 \geq 1 \]
Thus:
\[ x^2 \geq \frac{1}{2} \]
Taking the square root gives:
\[ |x| \geq \frac{1}{\sqrt{2}} \quad \text{or} \quad x \leq -\frac{1}{\sqrt{2}} \text{ or } x \geq \frac{1}{\sqrt{2}} \]

### Step 3: Combine the conditions
Now we have two conditions:
1. \( -1 \leq x \leq 1 \)
2. \( x \leq -\frac{1}{\sqrt{2}} \) or \( x \geq \frac{1}{\sqrt{2}} \)

We need to find the intersection of these two conditions.

- The interval \( -1 \leq x \leq 1 \) restricts \( x \) to values between -1 and 1.
- The condition \( x \leq -\frac{1}{\sqrt{2}} \) gives us the left part of the interval.
- The condition \( x \geq \frac{1}{\sqrt{2}} \) gives us the right part of the interval.

### Step 4: Determine the final domain
The values of \( x \) that satisfy both conditions are:
- From \( -1 \) to \( -\frac{1}{\sqrt{2}} \)
- From \( \frac{1}{\sqrt{2}} \) to \( 1 \)

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