Answer:
hey there hope this answer helps you out
Step-by-step explanation:
we have two functions f(x) and g(x) such that
[tex]f(x) = \frac{1}{x} [/tex]
and
[tex]g(x) = {x}^{2} + 6x[/tex]
solving for f ° g, we'll substitute the x in f(x) by the value of g(x)
[tex]fog = \frac{1}{g(x)} [/tex]
[tex]fog \: = \frac{1}{ {x}^{2} + 6x } [/tex]
taking x common in denominator
[tex]fog = \frac{1}{x(x + 6)} [/tex]
for a function to exist it should not have 0 in its denominator
checking the values of x for which the denominator of f ° g becomes 0 :-
so the function doesn't exist at values x = 0, -6
So, 0, -6 cannot be in the domain of f°g