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For each pair of functions [tex]f[/tex] and [tex]g[/tex] below, find [tex]f(g(x))[/tex] and [tex]g(f(x))[/tex]. Then, determine whether [tex]f[/tex] and [tex]g[/tex] are inverses of each other.

Simplify your answers as much as possible.
(Assume that your expressions are defined for all [tex]x[/tex] in the domain of the composition. You do not have to indicate the domain.)

(a) [tex]f(x) = \frac{1}{4x}, \, x \neq 0[/tex]
[tex]g(x) = \frac{1}{4x}, \, x \neq 0[/tex]
[tex]\[
\begin{array}{l}
f(g(x)) = \square \\
g(f(x)) = \square
\end{array}
\][/tex]

(b) [tex]f(x) = x + 6[/tex]
[tex]g(x) = x + 6[/tex]
[tex]\[
\begin{array}{l}
f(g(x)) = \square \\
g(f(x)) = \square
\end{array}
\][/tex]

Determine whether [tex]f[/tex] and [tex]g[/tex] are inverses of each other:
- [tex]f[/tex] and [tex]g[/tex] are inverses of each other.
- [tex]f[/tex] and [tex]g[/tex] are not inverses of each other.

Sagot :

GinnyAnswer
Let's solve the problem step-by-step for each pair of functions and determine if they are inverses.

### Pair 1:
Functions: [tex]\( f(x) = \frac{1}{4x} \)[/tex], [tex]\( g(x) = \frac{1}{4x} \)[/tex] where [tex]\( x \neq 0 \)[/tex].

1. Find [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{4x}\right) = \frac{1}{4 \left(\frac{1}{4x}\right)} = \frac{1}{4 \cdot \frac{1}{4x}} = \frac{1}{\frac{1}{x}} = x \][/tex]

2. Find [tex]\( g(f(x)) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{1}{4x}\right) = \frac{1}{4 \left(\frac{1}{4x}\right)} = \frac{1}{4 \cdot \frac{1}{4x}} = \frac{1}{\frac{1}{x}} = x \][/tex]

Since [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex], the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverses of each other for this pair.

### Pair 2:
Functions: [tex]\( f(x) = x + 6 \)[/tex], [tex]\( g(x) = x + 6 \)[/tex].

1. Find [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f(g(x)) = f(x + 6) = (x + 6) + 6 = x + 12 \][/tex]

2. Find [tex]\( g(f(x)) \)[/tex]:
[tex]\[ g(f(x)) = g(x + 6) = (x + 6) + 6 = x + 12 \][/tex]

Since [tex]\( f(g(x)) = x + 12 \)[/tex] and [tex]\( g(f(x)) = x + 12 \)[/tex], the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses of each other for this pair.

### Conclusion:
- For the first pair of functions [tex]\( f(x) = \frac{1}{4x} \)[/tex] and [tex]\( g(x) = \frac{1}{4x} \)[/tex]:
[tex]\[ f(g(x)) = x, \quad g(f(x)) = x \][/tex]
[tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverses of each other.

- For the second pair of functions [tex]\( f(x) = x + 6 \)[/tex] and [tex]\( g(x) = x + 6 \)[/tex]:
[tex]\[ f(g(x)) = x + 12, \quad g(f(x)) = x + 12 \][/tex]
[tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses of each other.
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