The domain of the function `f(x)=root(4)(x-sqrt(1-x^(2)))` is

Options

A. `[-1, (-1)/(sqrt2)]uu[(1)/(sqrt2),1]`
B. `[-1, 1]`
C. `(-oo, -(1)/(2)]uu[(1)/(sqrt2),oo)`
D. `[(1)/(sqrt2),1]`

Answer

To find the domain of the function \( f(x) = \sqrt[4]{x - \sqrt{1 - x^2}} \), we need to ensure that the expression inside the fourth root is non-negative, as well as ensuring that the square root is defined.

### Step 1: Ensure the square root is defined
The expression inside the square root, \( 1 - x^2 \), must be greater than or equal to zero:
\[ 1 - x^2 \geq 0 \]
This can be rearranged to:
\[ x^2 \leq 1 \]
Taking the square root of both sides gives:
\[ -1 \leq x \leq 1 \]

**Hint:** The square root function is defined only for non-negative values, so ensure the expression under the square root is non-negative.

### Step 2: Ensure the expression inside the fourth root is non-negative
Next, we need to ensure that the expression inside the fourth root, \( x - \sqrt{1 - x^2} \), is also non-negative:
\[ x - \sqrt{1 - x^2} \geq 0 \]
This can be rearranged to:
\[ x \geq \sqrt{1 - x^2} \]
Squaring both sides (noting that both sides are non-negative in the valid range of \( x \)):
\[ x^2 \geq 1 - x^2 \]
Rearranging gives:
\[ 2x^2 \geq 1 \]
\[ x^2 \geq \frac{1}{2} \]
Taking the square root of both sides gives:
\[ |x| \geq \frac{1}{\sqrt{2}} \]
This leads to two intervals:
\[ x \leq -\frac{1}{\sqrt{2}} \quad \text{or} \quad x \geq \frac{1}{\sqrt{2}} \]

**Hint:** When squaring both sides of an inequality, ensure that both sides are non-negative to avoid introducing extraneous solutions.

### Step 3: Combine the conditions
Now, we combine the results from Step 1 and Step 2. We have:
1. From Step 1: \( -1 \leq x \leq 1 \)
2. From Step 2: \( x \leq -\frac{1}{\sqrt{2}} \) or \( x \geq \frac{1}{\sqrt{2}} \)

Now we analyze the intervals:
- 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 \( \frac{1}{\sqrt{2}} \leq x \leq 1 \).

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

**Final Answer:** The domain of the function \( f(x) \) is \( [-1, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, 1] \).