Question

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

• A. $(-\infty, 0)$
• B. $(2, 5)$
• C. $(0, \infty)$
• D. $(-\infty, \infty)$

Hint:

An absolute value flips any negative input to its positive twin. So, if you feed a negative number into $x + |x|$, they will cancel out perfectly to zero and cause a division-by-zero crash. The function can only breathe when $x$ is strictly positive!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem requires us to find the set of all valid real numbers (the domain) for which the given radical fractional function $f(x)$ is mathematically defined.

Step 2: Detailed Explanation:
For a real-valued function containing an expression inside a square root in the denominator, two conditions must simultaneously hold true to avoid imaginary numbers and division-by-zero errors:

• The expression inside the square root must be non-negative: $x + |x| \ge 0$.

• The denominator cannot equal zero: $x + |x| \neq 0$.
Combining these two requirements, the function is valid if and only if: $$ x + |x| > 0 $$ Let's analyze the behavior of the absolute value function across different intervals for $x$:

• Case 1: When $x < 0$ (Negative numbers): By definition, if $x$ is negative, then $|x| = -x$. Substituting this into our expression gives: $$ x + (-x) = 0 $$ Since $0$ is not strictly greater than $0$, all negative numbers fail the requirement.

• Case 2: When $x = 0$: Substituting zero gives $0 + |0| = 0$, which also fails since the denominator becomes zero.

• Case 3: When $x > 0$ (Positive numbers): By definition, if $x$ is positive, then $|x| = x$. Substituting this into our expression gives: $$ x + x = 2x $$ Since $x > 0$, it follows that $2x > 0$, which perfectly satisfies our condition.
Therefore, the function is strictly defined only for all positive real numbers, which corresponds to the open interval $(0, \infty)$.

Step 3: Final Answer:
The domain of the function is $(0, \infty)$, matching option (C).