Domain of a Square Root Function

The function takes real values if the quantity under the radical satisfies the condition:

\[ x - 2 \ge 0 \]

Solving the inequality gives:

\[ x \ge 2 \]

Graphical Check: The graph "exists" only for $x$ values greater than or equal to 2.

The condition for real values is:

\[ |x - 1| \ge 0 \]

Because an absolute value expression is always greater than or equal to zero for all real numbers, the condition is satisfied for any $x$.

Domain: All real numbers $\mathbb{R}$, or $(-\infty, \infty)$.

Since division by zero is not allowed, the radicand must be strictly positive:

\[ x + 3 > 0 \]

Solving gives:

\[ x > -3 \]

Graphical Check: The graph starts just after $x = -3$ and extends to the right.

Two conditions must be met simultaneously:

1. Numerator: $x + 4 \ge 0 \Rightarrow x \ge -4$
2. Denominator: $x - 2 > 0 \Rightarrow x > 2$

The domain is the intersection of these sets:

\[ x > 2 \]

Condition: \[ \dfrac{x + 4}{x - 2} \ge 0 \]

The critical points are $x = -4$ and $x = 2$. Testing intervals:

• $(-\infty, -4]$: Expression is positive.
• $(-4, 2)$: Expression is negative.
• $(2, \infty)$: Expression is positive.

Domain: $(-\infty, -4] \cup (2, \infty)$. (Note: $x=2$ is excluded due to the denominator).

Condition: \[ -x^2 - 4 \ge 0 \Rightarrow x^2 + 4 \le 0 \]

Since $x^2 + 4$ is always at least 4 for any real $x$, it can never be less than or equal to 0.

Domain: Empty set (no real solutions).

Conditions:

• Numerator: $6 - x \ge 0 \Rightarrow x \le 6$
• Denominator: $x - 2 > 0 \Rightarrow x > 2$

Intersection: $2 < x \le 6$. In interval notation: $(2, 6]$.

Condition: \[ x^2 - 4 \ge 0 \Rightarrow (x - 2)(x + 2) \ge 0 \]

This inequality holds when $x$ is outside the roots $-2$ and $2$.

Domain: $(-\infty, -2] \cup [2, \infty)$.

Condition: \[ 4 - x^2 \ge 0 \Rightarrow (2 - x)(2 + x) \ge 0 \]

This inequality holds when $x$ is between the roots $-2$ and $2$.

Domain: $[-2, 2]$.