The given function is:

$F(x) = \frac{\sqrt{x+2}}{x-2}$

Domain:

The domain of a function is the set of all values of $x$ for which the function is defined.

1. For the square root: The expression inside the square root, $x + 2$, must be non-negative (since the square root of a negative number is not a real number). So we need: $x + 2 \geq 0 \implies x \geq -2$

2. For the denominator: The denominator $x - 2$ cannot be zero because division by zero is undefined. So we require: $x - 2 \neq 0 \implies x \neq 2$

Combining these two conditions, the domain of $F(x)$ is: $x \geq -2 \quad \text{and} \quad x \neq 2$ In interval notation, the domain is: $[-2, 2) \cup (2, \infty)$

Range:

The range is the set of all possible output values of $F(x)$. To determine the range, consider the behavior of the function:

• The numerator $\sqrt{x+2}$ gives non-negative values (starting from 0 when $x = -2$ and increasing as $x$ increases).
• The denominator $x - 2$ can be either positive or negative, depending on the value of $x$.

As $x \to -2^+$, $F(x) \to 0^+$ (since $\sqrt{x+2} \to 0$). As $x \to 2^-$, $F(x) \to -\infty$. As $x \to 2^+$, $F(x) \to +\infty$. As $x \to \infty$, $F(x) \to 0^+$.

Thus, the range of the function is all real numbers except 0, which the function approaches but never reaches: $(-\infty, 0) \cup (0, \infty)$

Would you like any additional details or have further questions?

Here are 5 related questions to explore this topic further:

1. What happens to the function $F(x)$ as $x \to 2$ from both sides?
2. How does the presence of the square root affect the graph of the function?
3. Can you sketch the graph of $F(x)$ based on its domain and range?
4. How do the domain and range change if the function were $F(x) = \frac{x+2}{x-2}$ without the square root?
5. How would the domain and range change if the denominator were $x^2 - 4$ instead?

Tip: Always check for both square roots and denominators when determining the domain, as they impose different restrictions on the function.