Question

Find the domain of the function. $$ f(x)=\frac{1}{\sqrt{|x|-x}} $$

Short answer

The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is \(x \geq 0\), or in interval notation: \([0, \infty)\).

Step-by-step solution

Identify the restrictions for the square root function.

The first restriction comes from the square root in the function: \(\sqrt{|x| - x}\). The value inside the square root must be non-negative, meaning: $$|x| - x \geq 0$$

Break down the absolute value function.

Since \(|x|\) is the absolute value of \(x\), we have two cases to consider: Case 1: \(x \geq 0\) In this case, \(|x| = x\), so: $$x - x \geq 0 \Rightarrow 0 \geq 0$$ The inequality is true for all \(x \geq 0\). Case 2: \(x < 0\) In this case, \(|x| = -x\), so: $$-x - x \geq 0 \Rightarrow 2x \leq 0 \Rightarrow x \leq 0$$ The inequality is true for all values of \(x \leq 0\).

Combine the cases and find the common solution.

From Step 2, we have two possible solutions: 1. \(x \geq 0\) 2. \(x \leq 0\) These inequalities define the whole real line, meaning the function is defined for all real numbers concerning the square root part.

Identify the restrictions for the denominator.

The second restriction comes from the denominator: \(\frac{1}{\sqrt{|x|-x}}\). It will be undefined when \(|x|-x = 0\). Solve for x: $$|x| - x = 0$$ We know that \(|x|-x\) is non-negative from Step 1 and Step 2, so: $$|x| = x$$ This equation is only true when \(x \geq 0\).

Determine the domain.

We now have two conditions for the domain of the function: 1. For the square root part, the function is defined for all real numbers: \(\mathbb{R}\). 2. For the denominator, the function is only defined when \(x \geq 0\). Taking both conditions into account, the domain of the function is the intersection of these sets: $$Domain(f(x)) = \{x \in \mathbb{R} : x \geq 0\}$$