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The known analytical formula for composite functions is:

f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x

The analytical formula for f [x] cannot be obtained through this method:

In[62]:= Clear["Global`*"]
RSolve[{f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x}, f[x], x] // FullSimplify

Out[63]= RSolve[{1/x + x == f[(1 + x)/Sqrt[x]]}, f[x], x]

and this method is also cannot

Clear["Global`*"]
Simplify /@ (f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x /. 
    Solve[y == Sqrt[x] + 1/Sqrt[x], x][[1]]) /. {y -> x, 
  Equal -> Rule}

enter image description here

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    $\begingroup$ Could you explain what The known analytical formula for composite functions is: f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x mean? According to Reduce, Reduce[Sqrt[x] + 1/Sqrt[x] == x + 1/x] is true when x=1 only. What is f role here? I am not reading your question right it seems. $\endgroup$
    – Nasser
    Jun 6 at 3:28
  • $\begingroup$ fF represents the corresponding relationship of the function $\endgroup$
    – csn899
    Jun 6 at 3:31
  • $\begingroup$ What is the analytical expression required for f [x]. Known conditions are f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x $\endgroup$
    – csn899
    Jun 6 at 3:32
  • $\begingroup$ Well, according to reduce, if you give me Sqrt[x] + 1/Sqrt[x] and want me to return back x + 1/x then only when x=1 this is possible. I am not sure what you mean by how to represent f in this case. Hopefully someone will have an answer. $\endgroup$
    – Nasser
    Jun 6 at 3:34

2 Answers 2

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Clear["Global`*"]

eqn1 = f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x;

sol = Solve[y == Sqrt[x] + 1/Sqrt[x], x] // Quiet

(* {{x -> 1/2 (-2 + y^2 - y Sqrt[-4 + y^2])}, 
    {x -> 1/2 (-2 + y^2 + y Sqrt[-4 + y^2])}} *)

Using first solution,

eqn2 = f[y] == Simplify[x + 1/x /. sol[[1]]] /.
  {y -> x, Equal -> Rule}

(* f[x] -> -2 + x^2 *)

Verifying,

eqn1 /. f -> Function[{x}, x^2 - 2] // Simplify

(* True *)

Using the second solution,

eqn3 = f[y] == Simplify[x + 1/x /. sol[[2]]] /.
  {y -> x, Equal -> Rule}

(* f[x] -> -2 + x^2 *)

This is identical to the first result.

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Assuming f is invertible:

eqn = f[y] == f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x;
Solve[Eliminate[eqn, {x}], f[y]]

InverseFunction::ifun: Inverse functions are being used. Values may be lost for multivalued inverses.

{{f[y] -> -2 + y^2}}

Note: The solution f is invertible on the implicit domain, which is the range of the argument Sqrt[x] + 1/Sqrt[x] (Re[y] >= 0). However, the method works on Solve[Eliminate[f[y] == f[Sqrt[x] + 1/Sqrt[x]] == 2, {x}], f[y]], even though f[y] -> 2 is not invertible. The warning arises because Solve uses InverseFunction[f][f[u]] to extract the argument u, which is what you see others (in the OP and other answer) doing by hand.

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