We are given two functions:

1. $f(x) = \frac{2 - x}{\sqrt{2x - 4}}$
2. $g(x) = \frac{x - 3}{x - 4}$

We need to determine the domain of the function $h(x) = f(x) + g(x)$.

Step 1: Find the domain of $f(x)$

The domain of $f(x)$ requires that:

• The expression inside the square root, $2x - 4$, must be greater than 0 because the square root function is defined for positive numbers. $2x - 4 > 0 \implies 2x > 4 \implies x > 2$
• Additionally, the denominator in $f(x)$ is $\sqrt{2x - 4}$, so $2x - 4 \neq 0$. This confirms that $x = 2$ is not included in the domain.

Thus, the domain of $f(x)$ is $x > 2$.

Step 2: Find the domain of $g(x)$

The domain of $g(x)$ requires that the denominator $x - 4$ is not zero: $x - 4 \neq 0 \implies x \neq 4$ Thus, $x \neq 4$.

Step 3: Combine the domain restrictions for $h(x) = f(x) + g(x)$

The domain of $h(x)$ will be the intersection of the domains of $f(x)$ and $g(x)$. The domain of $f(x)$ is $x > 2$, and the domain of $g(x)$ is all real numbers except $x = 4$.

Therefore, the domain of $h(x)$ is: $x > 2 \quad \text{and} \quad x \neq 4$

Final Answer:

The domain of $h(x) = f(x) + g(x)$ is $(2, 4) \cup (4, \infty)$.

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the range of the function $h(x) = f(x) + g(x)$?
2. How do we handle square roots when solving for domains?
3. What is the behavior of $h(x)$ near $x = 4$?
4. Can the function $f(x)$ or $g(x)$ have any discontinuities?
5. How would the domain change if the functions were composed as $f(g(x))$?

Tip: Always consider both the numerator and denominator when determining the domain of a rational function.