To find the domain of the composite function $f(g(x))$, where:

• $f(x) = \sqrt{8 - x}$
• $g(x) = \frac{1}{x - 3}$

Step 1: Domain of $f(x) = \sqrt{8 - x}$

For the square root function to be defined, the expression inside the square root must be non-negative. Thus, we require:

$8 - x \geq 0$

$x \leq 8$

So, the domain of $f(x)$ is $(-\infty, 8]$.

Step 2: Domain of $g(x) = \frac{1}{x - 3}$

The function $g(x)$ is undefined when the denominator is zero, so we exclude $x = 3$. Therefore, the domain of $g(x)$ is:

$x \in (-\infty, 3) \cup (3, \infty)$

Step 3: Domain of the Composite Function $f(g(x))$

Now, we evaluate the domain of $f(g(x))$. First, for $f(g(x))$ to be defined, $g(x)$ must lie in the domain of $f(x)$, which is $(-\infty, 8]$.

Since $g(x) = \frac{1}{x - 3}$, we require:

$g(x) = \frac{1}{x - 3} \leq 8$

Solving this inequality:

$\frac{1}{x - 3} \leq 8$

Multiply both sides by $(x - 3)^2$ (which is positive for $x \neq 3$):

$1 \leq 8(x - 3)$

$1 \leq 8x - 24$

$25 \leq 8x$

$x \geq \frac{25}{8}$

Thus, we require $x \geq \frac{25}{8}$ and $x \neq 3$ (since $g(x)$ is undefined at $x = 3$).

Final Answer: Domain of $f(g(x))$

The domain of the composite function $f(g(x))$ is:

$x \in \left[ \frac{25}{8}, 3 \right) \cup (3, \infty)$

Would you like more details on any part of the solution? Here are some related questions:

1. How do we determine the range of a composite function?
2. What happens when $f$ or $g$ are rational functions?
3. Can composite functions have a restricted range as well?
4. How do you simplify the inequality in a composite function's domain?
5. Why does the function $g(x) = \frac{1}{x - 3}$ create a discontinuity at $x = 3$?

Tip: When solving inequalities involving fractions, remember to consider both positive and negative cases separately if the expression contains a variable in the denominator.