assuming a typo, I see
f(x) = 1/(x-5)
g(x) = 6/x
so, f◦g = f(g) = 1/(g-5)
check for some value, say, x=2:
g(2) = 3
f(3) = -1/2
(f◦g)(2)=2/(6-5*2) = 2/-4 = -1/2
y-3 = e^(4-x)
ln(y-3) = 4-x
x = 4 - ln(y-3)
so, f^-1(x) = 4 - ln(x-3)
domain of f is all reals
range is y>3
asymptote is y=3
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