For each pair of functions $$ f $$ and $$ g $$ below, find $$ f(g(x)) $$ and $$ g(f(x)) $$. Then, determine whether $$ f $$ and $$ g $$ are inverses of each other. Simplify your answers as much as possible. (Assume that your expressions are defined for all $$ x $$ in the domain of the composition. You do not have to indicate the domain.) (a) $$ f(x)=-\frac{6}{x}, x \neq 0 $$ (b) $$ f(x)=2 x-1 $$ $$ \begin{array}{l} g(x)=-\frac{6}{x}, x \neq 0 \ f(g(x))=\square \ g(f(x))=\square \end{array} $$ $$ f $$ and $$ g $$ are inverses of each other $$ f $$ and $$ g $$ are not inverses of each other $$ \begin{array}{l} g(x)=2 x+1 \ f(g(x))=\square \ g(f(x))=\square \end{array} $$ $$ f $$ and $$ g $$ are inverses of each other $$ f $$ and $$ g $$ are not inverses of each other

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Let's solve the problem step by step for each pair of functions.

### Part (a)
Given:
- $$ f(x) = -\frac{6}{x} $$
- $$ g(x) = -\frac{6}{x} $$

#### Step 1: Find $$ f(g(x)) $$
We need to substitute $$ g(x) $$ into $$ f(x) $$:
$$
f(g(x)) = f\left(-\frac{6}{x}\right) = -\frac{6}{-\frac{6}{x}} = x
$$

#### Step 2: Find $$ g(f(x)) $$
Now substitute $$ f(x) $$ into $$ g(x) $$:
$$
g(f(x)) = g\left(-\frac{6}{x}\right) = -\frac{6}{-\frac{6}{x}} = x
$$

#### Conclusion for Part (a)
Since both compositions yield $$ x $$:
- $$ f(g(x)) = x $$
- $$ g(f(x)) = x $$

Thus, $$ f $$ and $$ g $$ are inverses of each other.

### Part (b)
Given:
- $$ f(x) = 2x - 1 $$
- $$ g(x) = 2x + 1 $$

#### Step 1: Find $$ f(g(x)) $$
Substituting $$ g(x) $$ into $$ f(x) $$:
$$
f(g(x)) = f(2x + 1) = 2(2x + 1) - 1 = 4x + 2 - 1 = 4x + 1
$$

#### Step 2: Find $$ g(f(x)) $$
Now substitute $$ f(x) $$ into $$ g(x) $$:
$$
g(f(x)) = g(2x - 1) = 2(2x - 1) + 1 = 4x - 2 + 1 = 4x - 1
$$

#### Conclusion for Part (b)
Since the compositions yield different results:
- $$ f(g(x)) = 4x + 1 $$
- $$ g(f(x)) = 4x - 1 $$

Thus, $$ f $$ and $$ g $$ are not inverses of each other.

### Summary of Results
For part (a):
$$
\begin{array}{l}
f(g(x)) = x \
g(f(x)) = x \
\text{Conclusion: } f \text{ and } g \text{ are inverses of each other.}
\end{array}
$$

For part (b):
$$
\begin{array}{l}
f(g(x)) = 4x + 1 \
g(f(x)) = 4x - 1 \
\text{Conclusion: } f \text{ and } g \text{ are not inverses of each other.}
\end{array}
$$
For part (a): $$ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x $$ $$ f $$ and $$ g $$ are inverses of each other. For part (b): $$ f(g(x)) = 4x + 1 \quad \text{and} \quad g(f(x)) = 4x - 1 $$ $$ f $$ and $$ g $$ are not inverses of each other.

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