How to solve the functional equation $3f(-x)+f(x)+f(\frac{1}{x})=x$

Question

I'm trying to solve this task:

Find all $f(x)$ satistying $3f(-x)+f(x)+f(\frac{1}{x})=x$

I tried to assume $x = 1$ and $x = -1$:

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

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

Solving this system gives:

$f(-1) = 1, f(1) = -1$

However, I'm not sure what to do next or if this would be helpful at all.

Could somebody please give a hint?

Thanks in advance.

Answer 1

We shall first prove that $f(x)$ is odd. The proof proceeds as follows:

Replacing $x$ with $-x$ in given equation & adding it to original equation, we get $$ \begin{align*} 4(f(x) + f(-x)) + \left( f\left( \frac{1}{x} \right) + f\left( -\frac{1}{x} \right) \right) &= 0\\ \implies 4 \left( f\left( \frac{1}{x} \right) + f\left( -\frac{1}{x} \right) \right) + (f(x) + f(-x)) &= 0 \tag{replace $x$ with $\tfrac{1}{x}$} \end{align*} $$

Hence, $f(x) + f(-x) = 0$, so $f(x)$ is odd.

Thus, $f\left( \tfrac{1}{x} \right) - 2 f(x) = x$ (as $3f(-x) = -3f(x)$)

Replacing $x$ with $\tfrac{1}{x}$ in this equation, we get: $$ \begin{align*} &\implies f(x) - 2 f\left( \frac{1}{x} \right) = \frac{1}{x}\\ &\implies f(x) - 2 (2 f(x) + x) = \frac{1}{x}\\ &\implies f(x) = -\frac{2x + \frac{1}{x}}{3} = -\frac{2x}{3}-\frac{1}{3x}\\ \end{align*} $$

Answer 2

Substituting $x\to \frac 1x$ in the original yields

$$3f\left(-\frac 1x\right)+f(x)+f\left(\frac{1}{x}\right)=\frac 1x \tag 1$$

I take the $$3f(-x)+f(x)+f\left(\frac{1}{x}\right)=x \tag 2$$

and I do $(2)-(1)$. We have:

$$3f(-x)-3f\left(-\frac 1x\right)=x-\frac 1x$$

Put $x\to -x$: $$3f(x)- 3f\left(\frac 1x\right)=-x+\frac{1}{x} \tag 3$$

$$f(x)- f\left(\frac 1x\right)=-\frac x3+\frac{1}{3x} \tag 4$$

Taking the original equation $3f(-x)+f(x)+f(\frac{1}{x})=x$ and sum the $(4)$ we have

$$2f(x)+3f(-x)=\frac 23x+\frac1{3x} \tag 5$$

Put again $x\to -x$ in the $(5)$

$$2f(-x)+3f(x)=-\frac 23x-\frac1{3x} \tag 6$$

Doing

$$3\cdot \left[2f(-x)+3f(x)=-\frac 23x-\frac1{3x}\right]-2\left[2f(x)+3f(-x)=\frac 23x+\frac1{3x}\right]$$ $$5f(x)=-2x-\frac 1x-\frac 43x-\frac{2}{3x}=-\frac{10}{3}x-\frac{5}{3x}$$

At the end

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

Answer 3

The method proposed by @Thomas Andrews shows that $f(x)$ must be a linear combination of $x, -x, \frac{1}{x},$ and $-\frac{1}{x}$, i.e. $$ f(x)=ax+\frac{b}{x}. \tag{1} $$ Plugging $(1)$ into the functional equation we obtain \begin{align} &3\left(-ax-\frac{b}{x}\right)+\left(ax+\frac{b}{x}\right)+\left(\frac{a}{x}+bx\right)=x \\ &\implies (-2a+b)x+\frac{a-2b}{x}=x \\ &\implies\begin{cases} -2a+b=1, \\ a-2b=0. \end{cases} \tag{2} \end{align} Solving $(2)$ we find $a=-\frac{2}{3}, b=-\frac{1}{3}$, hence $$ f(x)=-\frac{2x}{3}-\frac{1}{3x}. \tag{3} $$