I think some teachers out there must be misleading their students on this subject, because variations of that question keep coming up.
If a function has an inverse, then that inverse itself is a function. An inverse function is bound by the original function. Its domain is the range of the original function. If any of those values are excluded from the domain of the inverse, then it is not an inverse.
Also, not all functions have an inverse. In this case it would depend on the undeclared values of the constants in the function definition. For a function to have an inverse, it must be injective. For f(x), above, that would be true only if a = 0 or if some certain domain restriction were declared for f(x).
Followup:
The parameter assignments you introduced will make it easier to prove this point.
f(x) = (9x² + 5)/(2x)
Domain: x ≠ 0
Suppose that there is an f⁻¹(x), inverse of f(x). Then f⁻¹[f(x)] = x, for all x on the domain of f(x).
Let x = 1.
f(1) = 7
f⁻¹(7) = 1
Let x = 5/9.
f(5/9) = 7
f⁻¹(7) = 5/9
f⁻¹(7) = 1 and f⁻¹(7) = 5/9
Therefore, 1 = 5/9.
This contradiction is proof that f(x) has no inverse.
People often try to work around this arriving at invalid solutions. Here are a few I would expect:
f⁻¹(x) = [x ± √(x² - 45)]/9
This is not the inverse of a function, because it is not a function. It is two functions.
f⁻¹(x) = [x + √(x² - 45)]/9
No, because that would make f⁻¹[f(5/9)] = 1.
f⁻¹(x) = [x + √(x² - 45)]/9
No, because that would make f⁻¹[f(1)] = 5/9.
Change the domain of f⁻¹(x)?
The domain of f⁻¹(x) and the range of f(x) are (-∞, -√(45)] U [√(45), ∞). If any of those values were excluded from the domain of f⁻¹(x), then f(x) would be mapping numbers to values where the inverse is undefined.
Change the domain of f(x)?
It would be possible to restrict the domain of f(x) in such a way that it had an inverse. However, to change the domain is to change the definition of f(x). In that case we would simply be changing the question.