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We are given:

1. \( f(x) = 3x - 5 \)
2. \( (g \circ f)(x) = 2x + 1 \)

We need to determine \( g(x) \).

### Steps to solve:

- The composition \( (g \circ f)(x) \) means \( g(f(x)) \).
- Given \( f(x) = 3x - 5 \), substitute \( f(x) \) into the expression for \( g(f(x)) \):

\[
g(3x - 5) = 2x + 1
\]

Now, we let \( y = 3x - 5 \). This gives:

\[
g(y) = 2x + 1
\]

To express \( x \) in terms of \( y \):

\[
y = 3x - 5 \quad \Rightarrow \quad x = \frac{y + 5}{3}
\]

Now substitute \( x \) back into the expression \( g(y) = 2x + 1 \):

\[
g(y) = 2 \left( \frac{y + 5}{3} \right) + 1
\]
\[
g(y) = \frac{2(y + 5)}{3} + 1
\]
\[
g(y) = \frac{2y + 10}{3} + 1
\]
\[
g(y) = \frac{2y + 10}{3} + \frac{3}{3}
\]
\[
g(y) = \frac{2y + 13}{3}
\]

Thus, \( g(x) = \frac{2x + 13}{3} \).

### Answer:
The correct answer is **C: \( \frac{1}{3}(2x + 13) \)**.

Would you like any further clarification or a more detailed breakdown?

Here are five related questions:
1. What is the general method for solving composite functions?
2. How would this problem change if \( f(x) \) were a quadratic function?
3. Can you explain how to decompose complex functions into their simpler components?
4. How do you determine the domain of a composite function?
5. What is the significance of solving for \( g(x) \) in a real-world context?

**Tip**: Always double-check the composition of functions by substituting intermediate results into the original equations to ensure accuracy.

Jika f(x) = 3x - 5 dan (g°f)(x) = 2x + 1, maka g(x) = ...

Solve the composite function problem where f(x) = 3x - 5 and (g°f)(x) = 2x + 1 to find g(x). Includes detailed steps on function composition and substitution. Suitable for Grades 10-12.

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