Topic of the Question:
The topic of this question is set theory and functions, specifically focusing on finding the domain of a nested real-valued square root function.
Step 1 : Understanding the Question:
We are given the real-valued function $f(x) = \sqrt{x - \sqrt{1 - x^2}}$. We need to determine the set of all real values of $x$ for which this function is mathematically defined.
Step 2 : Key Formulas and Approach:
For a square root function of the form $\sqrt{g(x)}$ to yield real values, the expression inside the radical must be non-negative: $g(x) \geq 0$.
For a nested radical function, we apply this non-negativity constraint systematically from the innermost square root to the outermost square root.
The final domain of the function is determined by taking the intersection of all individual constraint intervals.
Step 3 : Detailed Explanation:
First, we look at the innermost radical expression, $\sqrt{1 - x^2}$. For this term to be defined, the expression inside must be non-negative: $1 - x^2 \geq 0$.
This simplifies to: $x^2 \leq 1 \implies -1 \leq x \leq 1$. This gives our first constraint interval: $x \in [-1, 1]$.
Next, we consider the outer radical, $\sqrt{x - \sqrt{1 - x^2}}$. For the outer square root to be defined, its entire inner expression must be non-negative: $x - \sqrt{1 - x^2} \geq 0 \implies x \geq \sqrt{1 - x^2}$.
Because the principal square root $\sqrt{1 - x^2}$ is always non-negative, the inequality $x \geq \sqrt{1 - x^2}$ requires $x$ to be non-negative as well: $x \geq 0$.
Combining the conditions $x \in [-1, 1]$ and $x \geq 0$, we restrict our working interval to $x \in [0, 1]$.
Since both sides of the inequality $x \geq \sqrt{1 - x^2}$ are non-negative on the interval $[0, 1]$, we can safely square both sides without introducing extraneous solutions: $x^2 \geq 1 - x^2$.
Rearranging this inequality gives: $2x^2 \geq 1 \implies x^2 \geq \frac{1}{2}$.
Taking the square root, and keeping in mind that $x \geq 0$, we obtain: $x \geq \frac{1}{\sqrt{2}}$. This gives our second constraint interval: $x \in \left[\frac{1}{\sqrt{2}}, \infty\right)$.
Finally, we find the intersection of the two constraint intervals: $[-1, 1] \cap \left[\frac{1}{\sqrt{2}}, \infty\right) = \left[\frac{1}{\sqrt{2}}, 1\right]$.
Step 4 : Final Answer:
The domain of the function is the interval $\left[\frac{1}{\sqrt{2}}, 1\right]$, which corresponds to Option (D).