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I tried the problem but i couldn't take above condition in question in terms of f(x)

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Are you sure this is related to Mathematica ? –  Sektor Sep 3 at 15:53

2 Answers 2

As suggested by george2079 the equation can be solved by substition:

Reverse@Simplify[f[x] == 2 x + 1/x - f[1/x]/2 /. f[1/x] -> 2/x + x - f[x]/2]
f[x] == 2 x   

To find $f^{-1}(4)$ you can use

f = 2 # &;
InverseFunction[f][4]
2
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readily derived.. f[x]==2x+1/x-f[1/x]/2, so f[1/x]==2/x+x-f[x]/2. Back substituting yields f[x]==2x+1/x-(2/x+x-f[x]/2)/2, which can be solved to yield f[x]==2x –  george2079 Sep 3 at 20:18

Rather than a guess I could provide a reasonable explanation (when lacking it usually leads astray) thus we are prompting another way offering also understanding.

Since the given equation is a functional one and Mathematica does not offer a direct functionality we have to deduce with the system an adequate scheme for solving such equations. Let's demonstrate it with Solve:

  1. we are solving related algebraic equations symbolically substituting appropriate function values with symbolic constants ( { f[x] == a, f[1/x] == b]})

  2. then we solve f[x] == 4

So we have to solve this system using a simple trick:

$$2f(x)+f(\frac{1}{x}) =4x+\frac{2}{x} $$and

$$2f(\frac{1}{x})+f(x) =\frac{4}{x}+2x $$

{ f[x] == a, f[1/x] == b} /. Solve[ 2 a + b == 4 x + 2/x && 2 b + a == 4/x + 2 x, {a, b}]
{{f[x] == 2 x, f[1/x] == 2/x}} 

now we have

Solve[ 2 x == 4, x]
{{x -> 2}}
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