Let `f(x) = 3 + x^2 + tan(pi/(2x)) , -1

Hello!

You wrote `f(x) = 3+x^2+tan(pi/(2x)).`

By the definition of an inverse function of `f(x),`  `f^(-1)(3)` is that number `x` for which `f(x) = 3.` Usually we require that such a number must be unique, otherwise `f^(-1)` would be a many-valued function.

a. In other words, we need to solve the equation `f(x) = 3.`

In our problem, `f(x)` takes any value infinitely many times, even at the given interval `(-1, 1),` even at...

Hello!

You wrote `f(x) = 3+x^2+tan(pi/(2x)).`

By the definition of an inverse function of `f(x),`  `f^(-1)(3)` is that number `x` for which `f(x) = 3.` Usually we require that such a number must be unique, otherwise `f^(-1)` would be a many-valued function.

a. In other words, we need to solve the equation `f(x) = 3.`

In our problem, `f(x)` takes any value infinitely many times, even at the given interval `(-1, 1),` even at any neighborhood of `x = 0.`

The cause of this is that `tan(pi/(2x))` tends to `+-oo` at points where `pi/(2x) = pi/2 + k pi` for some integer `k.` The `3+x^2` part remains finite and bounded at any finite interval and cannot prevent this behavior of `f(x).` These points are `x_k = 1/(1+2k)` and they tend to zero as `k` tends to `+-oo.`

Regardless of the number of solutions, the equation `f(x)=3,` which is equivalent to `x^2+tan(pi/(2x)) = 0,` cannot be solved exactly.

I might suppose that you misprint the formula, probably `f(x) = 3+x^2+tan(pi/2 x).` In that case, the only solution for `f(x)=3` at the interval `(-1,1)` is `x=0.` This is because `f` is strictly monotone on `(-1,1).` It is not obvious but true. Ask me if you need a proof.

b. If `f^(-1)(5)` exists, then by definition `f(f^(-1)(5)) = 5.`