$f( \frac {2x-1}{3} )$

I have a very creative solution to this problem that requires much less bashing. Let's examine the second equation (the one we have to solve). Let's first find $f(3)$ . If we want $f(\frac{2x-1}{3})=f(3)$ , then $3=\frac{2x-1}{3} \Longrightarrow x=5$ . Plugging this into the equation, we see that $f(3)=5$ (it's just a coincidence that plugging five into the equation produces five). Doing likewise for the other two functions ( $f(2)$ and $f(1)$ ) turns the final expression into $f(\frac{5-1}{-1}) \Longrightarrow f(-4)$ . We know how to tackle this, and doing so produces $\boxed{19}$ as the desired answer. Notice that we never even solved for the function! Great problem! :D

If we could solve for f(x) then it would be easy for us to solve the problem.

Let $f( \frac {2x-1}{3} )$ = $a( \frac {2x-1}{3} )^2$ + $b( \frac {2x-1}{3} )$ + $c$ . Then expanding it would yield:

$f( \frac {2x-1}{3} )$ = $a( \frac {4x^2 - 4x + 1}{9} )$ + $b( \frac {6x-3}{9} )$ + $\frac {9c}{9}$

$f( \frac {2x-1}{3} )$ = $\frac {4ax^2 - 4ax + a}{9}$ + $\frac {6bx - 3b}{9}$ + $\frac {9c}{9}$

$f( \frac {2x-1}{3} )$ = $\frac { 4ax^2 - 4ax + a + 6bx - 3b + 9c }{9}$

$f( \frac {2x-1}{3} )$ = $\frac {(4ax^2)+(-4ax + 6bx)+(a - 3b + 9c)}{9}$

Do not be confused about the grouping of terms; I just grouped the quadratic term, the linear terms, and the constant terms. If we compare this with $f( \frac {2x-1}{3} )$ = $\frac {4x^2 -10x -5}{9}$ , we can see that $4ax^2 = 4x^2$ , or $a =1$ (by cancelling $4x^2)$ ; $-4ax + 6bx = -10x$ , or $-4a + 6b = -10$ (by cancelling $x$ ) ; and $a - 3b + 9c = -5$ .

Now, we are left with the system

$a =1$

$-4a + 6b = -10$

$a - 3b + 9c = -5$

which, when solved, easily gives us $a = 1,$ $b = -1,$ and $c = -1$ . Substitute these values to $f( \frac {2x-1}{3} )$ = $a( \frac {2x-1}{3} )^2$ + $b( \frac {2x-1}{3} )$ + $c$ and you will obtain:

$f( \frac {2x-1}{3} )$ = $( \frac {2x-1}{3} )^2$ - $( \frac {2x-1}{3} )$ - $1$

Replace $\frac {2x-1}{3}$ with $x$ and you'll obtain: $f(x)$ = $x^2 - x - 1$

Now, $f( \frac { f(3) - f(2) }{f(1)} )$ = $f( \frac {(3^2 - 3 - 1) - (2^2 - 2 - 1)}{(1^2 - 1 - 1)} )$ = $f( \frac {5-1}{-1} )$ = $f(-4)$

Finally, $f(-4)$ = $(-4)^2 - (-4) - 1$ = $16 + 4 - 1$ = $\boxed{19}$

We can let $g(x) =\displaystyle\frac{2x-1}{3}$ and find for $g^{-1}(x)$ .

We can check that $g^{-1}(x) = \displaystyle\frac{3x+1}{2}$ .

Now we solve for $f(x)$ by computing for $\displaystyle f( g(\frac{3x+1}{2}))$ .

Why does this work? It is because in reality, you are solving for $\displaystyle f( g(g^{-1}(x))) = f(x)$ .

It is easy to verify that $f(x) = x^2 - x-1$

Then $f( \frac{f(3) - f(2)} {f(1)}) = f(\frac{(9-3-1) - (4-2-1)}{1-1-1}) = f(\frac{4}{-1}) = (-4)^2 -(-4) -1 = 16 + 4 - 1 = 19$

Replace x by (3x+1)/2 This will give f(x)

From $f( \frac{2x-1}{3})$ we need to find $f(x)$ .

Let's find $f(k)$ for some real number $k$ .

So, what should we put in place of $x$ to get $k$ ?
$\frac{2x-1}{3} =k \implies x= \frac{3k+1}{2}$

Thus, we obtain $f(k) = \frac { 4 ( \frac{3k+1}{2})^2 -10 (\frac{3k+1}{2}) -5}{9} \implies f(k) = k^2 -k-1 \implies f(x) = x^2-x-1$

So, we get $f(1)=-1, f(2)=1, f(3)=5$ .

Putting this values according to the question, we find $f(\frac{5-1}{-1}) =f(-4) =\boxed{19}$

By looking at given equation it can be concluded that f(x)=x^2-x-1. Then it becomes very simple. f(3)=5;f(2)=1;f(1)=-1.required answer is 19