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The question asks you to find \( f(g(x)) \) and \( g(f(x)) \), and determine whether the pair are inverses of each other. The given functions are:

\[
f(x) = 2x + 1
\]
\[
g(x) = \frac{x - 1}{2}
\]

### Finding \( f(g(x)) \):

To find \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \).

1. \( f(g(x)) = f\left( \frac{x - 1}{2} \right) \)
2. Using the definition of \( f(x) = 2x + 1 \), replace \( x \) with \( \frac{x - 1}{2} \):
\[
f\left( \frac{x - 1}{2} \right) = 2\left( \frac{x - 1}{2} \right) + 1
\]
3. Simplify:
\[
2\left( \frac{x - 1}{2} \right) = x - 1
\]
\[
f(g(x)) = (x - 1) + 1 = x
\]

Thus, \( f(g(x)) = x \).

### Finding \( g(f(x)) \):

Now find \( g(f(x)) \) by substituting \( f(x) \) into \( g(x) \).

1. \( g(f(x)) = g(2x + 1) \)
2. Using the definition of \( g(x) = \frac{x - 1}{2} \), replace \( x \) with \( 2x + 1 \):
\[
g(2x + 1) = \frac{(2x + 1) - 1}{2}
\]
3. Simplify:
\[
g(2x + 1) = \frac{2x}{2} = x
\]

Thus, \( g(f(x)) = x \).

### Conclusion:

Since both \( f(g(x)) = x \) and \( g(f(x)) = x \), the functions \( f(x) \) and \( g(x) \) are inverses of each other.

Would you like further details or have any questions?

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Here are 5 related questions:

1. How do you determine if two functions are inverses?
2. What is the graphical interpretation of inverse functions?
3. How can you verify if a function is one-to-one?
4. What happens if you calculate \( f(f(x)) \) or \( g(g(x)) \)?
5. Can inverse functions be determined algebraically for all types of functions?

**Tip:** Inverse functions undo each other's operations, meaning applying one after the other returns the original input.

Find f(g(x)) and g(f(x)) and determine whether the pair are inverses of each other. f(x) = 2x + 1 and g(x) = (x - 1) / 2.

This detailed solution explores how to find composite functions f(g(x)) and g(f(x)) for f(x) = 2x + 1 and g(x) = (x - 1) / 2, and verify if they are inverses of each other. Suitable for students in high school algebra.

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