Let's start with the given functions:
[tex]\[ f(x) = \frac{x-1}{2} \][/tex]
[tex]\[ g(x) = 2x + 1 \][/tex]
We need to find \( f(g(x)) \) and \( g(f(x)) \) to determine if \( f \) and \( g \) are inverses of each other.
### Step-by-Step Solution:
#### Calculating \( f(g(x)) \):
1. First, substitute \( g(x) = 2x + 1 \) into \( f(x) \):
[tex]\[
f(g(x)) = f(2x + 1)
\][/tex]
2. Replace \( x \) in \( f(x) \) with \( 2x + 1 \):
[tex]\[
f(2x + 1) = \frac{(2x + 1) - 1}{2}
\][/tex]
3. Simplify inside the numerator:
[tex]\[
f(2x + 1) = \frac{2x + 1 - 1}{2} = \frac{2x}{2}
\][/tex]
4. Simplify the fraction:
[tex]\[
f(2x + 1) = x
\][/tex]
So,
[tex]\[
f(g(x)) = x
\][/tex]
#### Calculating \( g(f(x)) \):
1. First, substitute \( f(x) = \frac{x-1}{2} \) into \( g(x) \):
[tex]\[
g(f(x)) = g\left( \frac{x-1}{2} \right)
\][/tex]
2. Replace \( x \) in \( g(x) \) with \( \frac{x-1}{2} \):
[tex]\[
g\left( \frac{x-1}{2} \right) = 2 \left( \frac{x-1}{2} \right) + 1
\][/tex]
3. Simplify inside the parentheses:
[tex]\[
g\left( \frac{x-1}{2} \right) = 2 \cdot \frac{x-1}{2} + 1 = x - 1 + 1
\][/tex]
4. Simplify the expression:
[tex]\[
g\left( \frac{x-1}{2} \right) = x
\][/tex]
So,
[tex]\[
g(f(x)) = x
\][/tex]
Since both compositions \( f(g(x)) \) and \( g(f(x)) \) simplify to \( x \), we conclude that:
[tex]\[
f \text{ and } g \text{ are inverses of each other.}
\][/tex]
### Final Answer:
[tex]\[
\boxed{x}
\][/tex]
[tex]\[
\boxed{x}
\][/tex]
\[ f \text{ and } g \text{ are inverses of each other.}
