Given the functions:
[tex]\[ f(x) = \frac{1}{x - 3} \][/tex]
[tex]\[ g(x) = \frac{6}{x} + 3 \][/tex]
### Finding \( f(g(x)) \)
Substitute \( g(x) \) into \( f(x) \):
[tex]\[ f(g(x)) = f\left( \frac{6}{x} + 3 \right) \][/tex]
Now, use the definition of \( f(x) \):
[tex]\[ f(g(x)) = \frac{1}{\left( \frac{6}{x} + 3 \right) - 3} \][/tex]
Simplify inside the denominator:
[tex]\[ f(g(x)) = \frac{1}{\frac{6}{x} + 3 - 3} \][/tex]
[tex]\[ f(g(x)) = \frac{1}{\frac{6}{x}} \][/tex]
[tex]\[ f(g(x)) = \frac{x}{6} \][/tex]
So, \( f(g(x)) \) simplifies to:
[tex]\[ f(g(x)) = \frac{x}{6} \][/tex]
### Finding \( g(f(x)) \)
Substitute \( f(x) \) into \( g(x) \):
[tex]\[ g(f(x)) = g\left( \frac{1}{x - 3} \right) \][/tex]
Now, use the definition of \( g(x) \):
[tex]\[ g(f(x)) = \frac{6}{\frac{1}{x - 3}} + 3 \][/tex]
Simplify the expression inside \( g(x) \):
[tex]\[ g(f(x)) = 6(x - 3) + 3 \][/tex]
Distribute and combine like terms:
[tex]\[ g(f(x)) = 6x - 18 + 3 \][/tex]
[tex]\[ g(f(x)) = 6x - 15 \][/tex]
So, \( g(f(x)) \) simplifies to:
[tex]\[ g(f(x)) = 6x - 15 \][/tex]
### Final Results:
Thus, we have:
[tex]\[ f(g(x)) = \frac{x}{6} \][/tex]
[tex]\[ g(f(x)) = 6x - 15 \][/tex]
