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

Options

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

Answer

To find the domain of the function \( f(x) = \sqrt{x - \sqrt{1 - x^2}} \), we need to ensure that the expression inside the square root is non-negative. This requires two conditions to be satisfied:

1. The expression \( 1 - x^2 \) must be non-negative.
2. The expression \( x - \sqrt{1 - x^2} \) must also be non-negative.

### Step 1: Determine when \( 1 - x^2 \geq 0 \)

We start with the inequality:
\[ 1 - x^2 \geq 0 \]
Rearranging gives:
\[ x^2 \leq 1 \]
This can be factored as:
\[ (x - 1)(x + 1) \leq 0 \]
To solve this inequality, we find the critical points, which are \( x = -1 \) and \( x = 1 \). We can test intervals around these points:

- For \( x < -1 \) (e.g., \( x = -2 \)): \( (-2 - 1)(-2 + 1) = (-3)(-1) = 3 > 0 \) (not in the solution).
- For \( -1 < x < 1 \) (e.g., \( x = 0 \)): \( (0 - 1)(0 + 1) = (-1)(1) = -1 < 0 \) (in the solution).
- For \( x > 1 \) (e.g., \( x = 2 \)): \( (2 - 1)(2 + 1) = (1)(3) = 3 > 0 \) (not in the solution).

Thus, the solution to this inequality is:
\[ -1 \leq x \leq 1 \]

### Step 2: Determine when \( x - \sqrt{1 - x^2} \geq 0 \)

Next, we analyze the second condition:
\[ x - \sqrt{1 - x^2} \geq 0 \]
Rearranging gives:
\[ x \geq \sqrt{1 - x^2} \]
Squaring both sides (noting that both sides are non-negative in the valid range) gives:
\[ x^2 \geq 1 - x^2 \]
This simplifies to:
\[ 2x^2 \geq 1 \quad \Rightarrow \quad x^2 \geq \frac{1}{2} \]
Taking square roots gives:
\[ |x| \geq \frac{1}{\sqrt{2}} \quad \Rightarrow \quad x \leq -\frac{1}{\sqrt{2}} \quad \text{or} \quad x \geq \frac{1}{\sqrt{2}} \]

### Step 3: Combine the conditions

Now we need to find the intersection of the two conditions:
1. From \( -1 \leq x \leq 1 \)
2. From \( x \leq -\frac{1}{\sqrt{2}} \) or \( x \geq \frac{1}{\sqrt{2}} \)

The critical points are \( -1 \), \( -\frac{1}{\sqrt{2}} \), \( \frac{1}{\sqrt{2}} \), and \( 1 \).

- The interval \( -1 \leq x \leq 1 \) intersects with \( x \leq -\frac{1}{\sqrt{2}} \) to give \( -1 \leq x \leq -\frac{1}{\sqrt{2}} \).
- The interval \( -1 \leq x \leq 1 \) intersects with \( x \geq \frac{1}{\sqrt{2}} \) to give \( \frac{1}{\sqrt{2}} \leq x \leq 1 \).

### Final Domain

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