Question

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

• A. \( \left[-1, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, 1\right] \)
• B. \( [-1, 1] \)
• C. \( \left(-\infty, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, +\infty\right] \)
• D. \( \left[\frac{1}{\sqrt{2}}, 1\right] \)

Hint:

Always analyze the sign profiles before squaring an inequality. In this problem, the condition \(x \geq \sqrt{1-x^2}\) immediately eliminates all negative numbers from the domain, allowing you to instantly discard options (1), (2), and (3) without completing any further algebraic calculations!

Answer 1

The Correct Option is D

Concept: For a real-valued function, the domain is the set of all real numbers \(x\) for which the function expression is mathematically well-defined. For square root functions of the form \(\sqrt{g(x)}\), the expression inside the square root must be non-negative: \[ g(x) \geq 0 \] When multiple square roots are nested, these constraints must be applied systematically to all radical expressions from the inside out, and the final domain is the intersection of all resulting solution sets. Step 1: Applying the condition for the inner square root.
Consider the inner square root expression \(\sqrt{1 - x^2}\). For this term to be defined: \[ 1 - x^2 \geq 0 \quad \Rightarrow \quad x^2 \leq 1 \] Taking the square root on both sides defines our first constraint interval: \[ -1 \leq x \leq 1 \quad \Rightarrow \quad x \in [-1, 1] \quad \cdots (1) \]

Step 2: Applying the condition for the outer square root.
For the outer square root to be well-defined, its entire internal expression must be non-negative: \[ x - \sqrt{1 - x^2} \geq 0 \quad \Rightarrow \quad x \geq \sqrt{1 - x^2} \quad \cdots (2) \] Since a principal square root value is always non-negative (\(\sqrt{1 - x^2} \geq 0\)), equation (2) dynamically forces \(x\) to also be non-negative: \[ x \geq 0 \quad \cdots (3) \] Combining constraint (1) and condition (3) yields a narrowed temporary boundary: \(x \in [0, 1]\). Now, because both sides of equation (2) are non-negative in this interval, we can safely square both sides without introducing extraneous solution anomalies: \[ x^2 \geq 1 - x^2 \] \[ 2x^2 \geq 1 \quad \Rightarrow \quad x^2 \geq \frac{1}{2} \] Since we established \(x \geq 0\), taking the positive square root gives: \[ x \geq \frac{1}{\sqrt{2}} \quad \cdots (4) \]

Step 3: Finding the intersection of all constraints.
We find the final domain by taking the intersection of all structural constraint intervals:
• From inner radical: \(x \in [-1, 1]\)
• From outer radical: \(x \in \left[\frac{1}{\sqrt{2}}, +\infty\right)\) \[ \text{Domain} = [-1, 1] \cap \left[\frac{1}{\sqrt{2}}, +\infty\right) = \left[\frac{1}{\sqrt{2}}, 1\right] \]