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Simplify your answers as much as possible. (Assume that your expressions are defined. You do not have to indicate the domain.)

(a)
[tex]\[
\begin{array}{l}
f(x) = \frac{x-1}{2} \\
g(x) = 2x + 1 \\
f(g(x)) = \square \\
g(f(x)) = \square
\end{array}
\][/tex]

Are \( f \) and \( g \) inverses of each other?

A. \( f \) and \( g \) are inverses of each other.
B. [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses of each other.



Answer :

GinnyAnswer
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.}

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