If $f(1-x) + 2f(x) = 3x$, what is $f(0)?$

Question

I did the following:

$$f(1-0) + 2f(0) = 3\cdot 0$$

$$f(1) + 2f(0) = 0$$

This reminds me of the equation of the straight line in the plane, then:

$$\left< \begin{pmatrix} {1}\\ {2} \end{pmatrix} , \begin{pmatrix} {f(1)}\\ {f(0)} \end{pmatrix} \right> =0$$

$(1,2)$ is a normal vector to $(f(1),f(0))$, then It's possible that: $(f(1)=-2,f(0)=1)$ because:

$$\left< \begin{pmatrix} {1}\\ {2} \end{pmatrix} , \begin{pmatrix} {-2}\\ {1} \end{pmatrix}\right> =-2+2=0$$

With generality, It is possible that for all $\alpha$:

$$\left< \begin{pmatrix} {1}\\ {2} \end{pmatrix} , \begin{pmatrix} {\alpha\cdot (-2)}\\ {\alpha \cdot 1} \end{pmatrix}\right> = \left< \begin{pmatrix} {1}\\ {2} \end{pmatrix} , \alpha \begin{pmatrix} { -2}\\ { 1} \end{pmatrix}\right> =1\cdot -2\alpha+2\alpha \cdot 1=-2\alpha + 2 \alpha =0$$

I tagged with "functional equations" because it seems to be related. I don't know if my solution is correct, what I'm saying with this is that there are infinite solutions but I don't know if assuming arbitrary values for $f(1),f(0)$ can be made because of $3x$.

EDIT: I know that It is possible to solve: $$f(1) + 2f(0) = 0$$ $$f(0) + 2f(1) = 3$$

And get the solution. But why do I have to use the latter instead of the former?

Answer 1

plugging in 0, you get $$ f(1) + 2f(0) = 0 $$ Now plug in 1 to get $$ f(0) + 2f(1) = 3 $$ and solve the linear system.

Answer 2

Just in case if you want to solve for the actual function next time:
Substituting $x$ with $1-x$ $$f(1-(1-x))+2f(1-x)=3(1-x)$$$$f(x)+2f(1-x)=3-3x\tag{1}$$Now multiply the original equation by 2:$$2f(1-x)+4f(x)=6x\tag{2}$$ Subtract: $(2)-(1)$ We get $$3f(x)=9x-3$$Therefore$$f(x)=3x-1$$$$f(0)=-1$$

Answer 3

$$f(1-x)+2f(x)=3x$$ set $1-x=t$ $$f(t)+2f(1-t)=3(1-t)$$

so $f(1-x)+2f(x)=3x$ and $f(x)+2f(1-x)=3(1-x)$ $$f(x)=3x-1$$ $f(0)=-1$

Answer 4

Another approach could be this:

$f(1-x) + 2f(x) = 3x$

$f(1-(1-x))+2f(1-x) = 3(1-x) \implies 2f(1-x) + f(x) = 3-3x$

thus we have a system

$$\left(\begin{array}{c c} 1 & 2 \\ 2 & 1 \end{array}\right)\left(\begin{array}{c} f(1-x) \\ f(x) \end{array}\right) = \left(\begin{array}{c} 3x \\ 3-3x \end{array}\right)$$ which has unique solution $f(x) = 3x - 1$, $f(1-x) = 2-3x$, and indeed, quick check shows that this is consistent.