Question

The domain of the function $f(x) = \sqrt{x - 1 + \sqrt{6 - x}}$ is

• A. $[1, \infty)$
• B. $[1, 6]$
• C. $(-\infty, 6)$
• D. $(-\infty, 6]$

Hint:

You can test the boundary points of the options to find the answer immediately! Notice what happens if you pick an easy value outside the intervals, like $x = 0$ (present in options C and D): $f(0) = \sqrt{-1} + \sqrt{6}$, which contains an imaginary number. This rules out (C) and (D) instantly. Testing a large value like $x = 7$ rules out (A), leaving (B) as the only mathematically sound choice.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the mathematical domain of a real-valued composite function composed of the sum of two distinct square-root functions.

Step 2: Key Formula or Approach:
For a real-valued square root function $\sqrt{g(x)}$ to be properly defined, the mathematical expression inside the radical must be non-negative: $$g(x) \ge 0$$ Since our function is a combination of two terms, the overall domain is the intersection of the individual domains of both terms: $$\text{Domain} = \text{Domain}_1 \cap \text{Domain}_2$$

Step 3: Detailed Explanation:
Let's analyze the first square root term, $\sqrt{x - 1}$: $$x - 1 \ge 0 \implies x \ge 1 \implies x \in [1, \infty)$$ Now let's analyze the second square root term, $\sqrt{6 - x}$: $$6 - x \ge 0 \implies 6 \ge x \implies x \le 6 \implies x \in (-\infty, 6]$$ To find the final valid domain where both functions are simultaneously active, take the intersection of both intervals: $$\text{Domain} = [1, \infty) \cap (-\infty, 6]$$ This combination spans all real numbers starting from $1$ up to and including $6$: $$\text{Domain} = [1, 6]$$

Step 4: Final Answer:
The domain of the function is the closed interval $[1, 6]$, which corresponds to option (B).