Question

The domain of the function \(f(x) = \sqrt{\frac{x}{1+x}}\) is \(\ldots\)

• A. \((-\infty ,-1)\cup [0,\infty )\)
• B. \((-\infty ,-1]\cup [0,\infty )\)
• C. \((-\infty ,-1)\cap [0,\infty )\)
• D. all R

Hint:

Need x/(1+x) >= 0 and x != -1. Use the sign chart.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The square root is defined when its argument is not negative, and the fraction needs a nonzero denominator. So we need \(\dfrac{x}{1+x} \ge 0\) with \(x \ne -1\).

Step 2: Key Formula or Approach:
A fraction is non-negative when the numerator and the denominator have the same sign, or the numerator is zero. The critical points are \(x = 0\) (zero of numerator) and \(x = -1\) (zero of denominator).

Step 3: Detailed Explanation:
Check each interval:
For \(x < -1\): numerator negative, denominator negative, so the fraction is positive. Allowed.
For \(-1 < x < 0\): numerator negative, denominator positive, so the fraction is negative. Not allowed.
For \(x = 0\): the fraction is 0, and \(\sqrt{0}\) is defined. Allowed.
For \(x > 0\): both positive, so the fraction is positive. Allowed.
At \(x = -1\) the function is undefined, so -1 is excluded and 0 is included:
\[ (-\infty, -1) \cup [0, \infty) \]
Option (B) wrongly includes \(-1\). Option (C) uses an intersection, which is empty. Option (D) ignores the restriction.

Final Answer:
The domain is \((-\infty,-1)\cup[0,\infty)\), option (A). \[ \boxed{(-\infty,-1)\cup[0,\infty) \text{ (A)}} \]