Write a formula for f^-1 (x) if f(x) = (x+5)/(3x-4) and state the domain of f-1
y=(x+5)/(3x-4)
3xy - 4y = x + 5
x(3y-1) = 4y+5
x = (4y+5)/(3y-1)
so, f^-1(x) = (4x+5)/(3x-1)
The domain is all reals except x = 1/3
Note that 1/3 is the horizontal asymptote of f(x). Since f cannot take on the value of 1/3, that value is excluded from the domain of f^-1(x)