Let's solve for both $f(g(x))$ and $g(f(x))$ step by step.

1. Solving $f(g(x))$:

The function $f(x) = 3x + 5$ and $g(x) = \frac{x}{3} - 5$.

To find $f(g(x))$, substitute $g(x)$ into $f(x)$:

$f(g(x)) = f\left( \frac{x}{3} - 5 \right)$

Now apply the function $f(x) = 3x + 5$:

$f(g(x)) = 3\left( \frac{x}{3} - 5 \right) + 5$

Distribute the 3:

$f(g(x)) = 3 \times \frac{x}{3} - 3 \times 5 + 5 = x - 15 + 5$

Simplify:

$f(g(x)) = x - 10$

2. Solving $g(f(x))$:

Now, we want to find $g(f(x))$, where $g(x) = \frac{x}{3} - 5$ and $f(x) = 3x + 5$.

Substitute $f(x)$ into $g(x)$:

$g(f(x)) = g(3x + 5)$

Now apply the function $g(x) = \frac{x}{3} - 5$:

$g(f(x)) = \frac{3x + 5}{3} - 5$

Simplify:

$g(f(x)) = \frac{3x}{3} + \frac{5}{3} - 5 = x + \frac{5}{3} - 5$

Combine terms:

$g(f(x)) = x + \frac{5}{3} - \frac{15}{3} = x - \frac{10}{3}$

Final Results:

• $f(g(x)) = x - 10$
• $g(f(x)) = x - \frac{10}{3}$

Would you like more details on any step or further clarifications?

Here are 5 related questions to explore:

1. How would the result change if $f(x)$ were a quadratic function?
2. Can you compute $f(h(x))$ and $h(f(x))$?
3. What happens if you compose $f(x)$ with itself, i.e., $f(f(x))$?
4. What are the inverses of $f(x)$ and $g(x)$, and can you verify the compositions using them?
5. How would you approach composing three functions, say $f(g(h(x)))$?

Tip: Function composition can be thought of as feeding the output of one function as the input to another. Understanding the inner workings of each function helps simplify the overall expression.