For any non-zero real number $$ x $$, let
$$ f(x) + 2f\left(\frac{1}{x}\right) = 3x. $$
Then, the sum of all possible values of $$ x $$ for which $$ f(x) = 3 $$ is:

Question

For any non-zero real number $$ x $$, let  
$$ f(x) + 2f\left(\frac{1}{x}\right) = 3x. $$  
Then, the sum of all possible values of $$ x $$ for which $$ f(x) = 3 $$ is:

Accepted Answer

1. **Substitute $$ x = \frac{1}{x} $$ into the equation:**  
   - Original equation:  
     $$ f(x) + 2f\left(\frac{1}{x}ight) = 3x. $$  
   - Substituting $$ x = \frac{1}{x} $$ gives:  
     $$ f\left(\frac{1}{x}ight) + 2f(x) = \frac{3}{x}. $$

2. **Solve the system of equations:**  
   - Multiply the second equation by 2:  
     $$ 2f\left(\frac{1}{x}ight) + 4f(x) = \frac{6}{x}. $$  
   - Subtract the original equation from this result:  
     $$
     [2f\left(\frac{1}{x}ight) + 4f(x)] - [f(x) + 2f\left(\frac{1}{x}ight)] = \frac{6}{x} - 3x.
     $$
     Simplify:  
     $$
     3f(x) = \frac{6}{x} - 3x.
     $$
   - Solve for $$ f(x) $$:  
     $$
     f(x) = \frac{2}{x} - x.
     $$

3. **Set $$ f(x) = 3 $$ and solve for $$ x $$:**  
   - Substitute into the equation:  
     $$
     \frac{2}{x} - x = 3.
     $$
   - Rearrange terms to form a quadratic equation:  
     $$
     x^2 + 3x - 2 = 0.
     $$
   - Factorize:  
     $$
     (x + 2)(x + 1) = 0.
     $$
   - Roots are:  
     $$ x = -2, \, x = -1. $$

4. **Sum of roots:**  
   Using the formula for the sum of roots of a quadratic equation ($$ -b/a $$):  
   $$
   	ext{Sum of roots} = (-3)/1 = -3.
   $$

Thus, the sum of all possible values of $$ x $$ for which $$ f(x) = 3 $$ is **-3**.

Answer

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