We are given two functions:

• $f(x) = 3x^{-1} = \frac{3}{x}$
• $g(x) = 4x + 8$

Let's find each composition step by step.

(a) $f(g(x))$

This means plugging $g(x)$ into $f(x)$. We have $g(x) = 4x + 8$, so:

$f(g(x)) = f(4x + 8) = \frac{3}{4x + 8}$

Thus, $f(g(x)) = \frac{3}{4x + 8}$.

(b) $g(f(x))$

This means plugging $f(x)$ into $g(x)$. We have $f(x) = \frac{3}{x}$, so:

$g(f(x)) = g\left(\frac{3}{x}\right) = 4\left(\frac{3}{x}\right) + 8 = \frac{12}{x} + 8$

Thus, $g(f(x)) = \frac{12}{x} + 8$.

(c) $f(f(x))$

This means plugging $f(x)$ into itself. We know $f(x) = \frac{3}{x}$, so:

$f(f(x)) = f\left(\frac{3}{x}\right) = \frac{3}{\frac{3}{x}} = \frac{3x}{3} = x$

Thus, $f(f(x)) = x$.

---

Domain Analysis

Now, let's determine the domain for each composition function:

• (a) $f(g(x)) = \frac{3}{4x + 8}$: The expression is undefined when the denominator equals zero, i.e., $4x + 8 = 0$, which gives $x = -2$. Therefore, the domain of $f(g(x))$ is all real numbers except $x = -2$.

• (b) $g(f(x)) = \frac{12}{x} + 8$: This expression is undefined when $x = 0$ (division by zero). Therefore, the domain of $g(f(x))$ is all real numbers except $x = 0$.

• (c) $f(f(x)) = x$: The domain of $f(x)$ is $x \neq 0$ (since division by zero is undefined in $f(x)$). Thus, the domain of $f(f(x))$ is all real numbers except $x = 0$.

---

Final Answers

(a) $f(g(x)) = \frac{3}{4x + 8}$
(b) $g(f(x)) = \frac{12}{x} + 8$
(c) $f(f(x)) = x$

Domains:

• (a) Domain: All real numbers except $x = -2$
• (b) Domain: All real numbers except $x = 0$
• (c) Domain: All real numbers except $x = 0$

---

Would you like further details or have any questions?

---

Relative Questions:

1. How do you determine the domain of a function?
2. What happens when you compose functions with restricted domains?
3. Can you explain how the composition of functions works step by step?
4. How would the domain of $f(g(x))$ change if $g(x)$ was a different function?
5. What is the difference between $f(g(x))$ and $g(f(x))$ in terms of application?

Tip:

When finding the domain of composite functions, always check where the individual functions are undefined.