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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?

Red Banana
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  • What if you plugged in $x=1$ and got an expression for $f(1)$ which you could plug into the expression when $x=0$. – Dave Jun 09 '16 at 21:10
  • @Dave See the edit. – Red Banana Jun 09 '16 at 21:12
  • oh, hadn't seen that. – Dave Jun 09 '16 at 21:13
  • Why? Because $f(1) + 2f(0) = 0$ is just one equation with two unknowns. There are infinitely many solutions that fit this. All you have shown is that there is a particular relation between the two values $f(1)$ and $f(0)$. However there are more information in the equation you started with and taking $x=1$ gives you enough to close the equation system and get a unique solution out. – Winther Jun 09 '16 at 21:22
  • @Winther Yes. Some minutes after "solving" it, I felt that the $3x$ could "lock" the equation, but didn't know how to express it. – Red Banana Jun 09 '16 at 21:25

4 Answers4

10

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$$f(x)=3x-1$$$$f(0)=-1$$

Surb
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Zack
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  • I took the liberty to improve the equation labelling of your answer (with the command \tag{ }). Feel free to rollback the edit if you don't like it. – Surb Jun 09 '16 at 21:23
6

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.

gt6989b
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  • I am aware of this method. But why does it need to be this method? – Red Banana Jun 09 '16 at 21:07
  • @Voyska i tend to use the simplest mechanism available at hand. Not sure about other methods, but this would seem like it works ok – gt6989b Jun 09 '16 at 21:08
  • Yes, but the problem is that my method also "works". – Red Banana Jun 09 '16 at 21:09
  • @Voyska this proves conclusively that there is a unique solution to the problem (percisely because of $3x$ as you indicated) not an infinite amount... – gt6989b Jun 09 '16 at 21:10
  • @Voyska your method produces only one constraint for the problem. Any other solution on the same line, according to you, will work ok. But this is not true. – gt6989b Jun 09 '16 at 21:10
2

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

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

Mike Meng
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2

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.

Ennar
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