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.
