Functional equation to determine $f(-1)$ 
There's a function $f$ satisfying: 

$$f\left(\frac{1}{1−x}\right)+2\cdot f\left(\frac{x−1}{x}\right)=3x$$

Find the value of $f(−1)$.

I have no idea how to solve this one. If anyone could help me to understand it, I would be grateful.
 A: Plug in $x=2$. Then you obtain $f(-1)+2f(1/2)=6$. On the other hand, for $x=-1$, you obtain $f(1/2)+2f(2)=-3$. For $x=1/2$, you get $f(2)+2f(-1)=3/2$. To sum up, $$f(-1)+2f(1/2)= 6\\ f(1/2)+2f(2)=-3 \\ f(2)+2f(-1)=3/2$$
Now take the values $f(-1),f(1/2),f(2)$ as unknowns and solve the linear system of equations.
A: The usual approach in these types of questions is just to fiddle around a bit. So, let's see how we can find $f(-1)$. To get this value, we have two options: choose an $x$ such that $1/(1-x)=-1$ or $(x-1)/x=-1$. For the first option, we get $x=2$ and the second option we get $x=1/2$. Now, let's fill this in and we get for $x=2$:
$$ f(-1) + 2\cdot f(1/2) = 6 $$
and for $x=1/2$
$$ f(2) + 2\cdot f(-1) = 3/2. $$
We don't know $f(1/2)$ and $f(2)$. So, let's see what we can say about those. For $f(2)$, we use the value of $x=-1$ to get
$$ f(1/2) + 2\cdot f(2) = -3. $$
This is an incredibly nice result since now we have $3$ unknowns and three equations which we should be able to solve using linear algebra. In this case you get $f(-1)=2$, $f(1/2)=2$ and $f(2)=-5/2$.
A: First let $x=2$, so $$f(-1)+2f(\frac{1}{2})=6$$
Similarly, let $x=-1$, $$f(\frac{1}{2})+2f(2)=-3$$
and let $x=\frac{1}{2}$, $$f(2)+2f(-1)=\frac{3}{2}$$
Solve this problem becomes solving the three simultaneous equations.
A: set $x = -1$ into to the formula of $f$:
$$f\left(\frac{1}{1−x}\right)+2\cdot f\left(\frac{x−1}{x}\right)=3x $$
you'll have for $x = -1$:
$$f\left(\frac{1}{1−(-1)}\right)+2\cdot f\left(\frac{(-1)−1}{(-1)}\right)=-3$$
$$ = f\left(\frac{1}{2}\right)+2\cdot f\left(2\right)=-3$$
Notice that the denominator of is different then $0$
and for $x = \frac{1}{2}$:
$$f\left(2\right)+2\cdot f\left(-1\right)=1.5 $$
 for $x=2$:
$$f(-1)+2\cdot f\left(\frac{1}{2} \right)= 6 $$
Then: 
$f(-1) = 6 - 2f\left(\frac{1}{2} \right) = \frac{-f(2)}{2} + \frac{3}{4}$ ,
 as $ \ -2f\left(\frac{1}{2} \right) = 4f(2) + 6 $ 
Therefore $f(2) = -2.5 \ \Longrightarrow \ f(-1) = \frac{-2.5}{-2} + \frac{3}{4} = 2$ 
A: Substitute $1-\frac{1}{x}$ instead of $x$.
We obtain:
$$f\left(\frac{1}{1-1+\frac{1}{x}}\right)+2f\left(1-\frac{1}{1-\frac{1}{x}}\right)=3\left(1-\frac{1}{x}\right)$$ or
$$f(x)+2f\left(\frac{1}{1-x}\right)=3\left(1-\frac{1}{x}\right).$$
Now, substitute $1-\frac{1}{x}$ instead of $x$ in the last equation.
We obtain:
$$f\left(1-\frac{1}{x}\right)+2f\left(\frac{1}{1-1+\frac{1}{x}}\right)=3\left(1-\frac{1}{1-\frac{1}{x}}\right)$$ or
$$f\left(1-\frac{1}{x}\right)+2f(x)=\frac{3}{1-x}.$$
Id est,
$$f(x)=3\left(1-\frac{1}{x}\right)-2f\left(\frac{1}{1-x}\right)=\frac{3(x-1)}{x}-2\left(3x-2f\left(1-\frac{1}{x}\right)\right)=$$
$$=\frac{3(x-1)}{x}-2\left(3x-2\left(\frac{3}{1-x}-2f(x)\right)\right),$$
which gives
$$9f(x)=\frac{3(x-1)}{x}-6x+\frac{12}{1-x}$$ or
$$f(x)=\frac{x-1}{3x}-\frac{2x}{3}+\frac{4}{3(1-x)}$$ and $$f(-1)=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}=2.$$
