Solve 2f(x)+f(1/x)=1/x

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

Solve for f (complex solution)

$$f=\frac{1}{2x^{2}+1}$$

$$x\neq -\frac{\sqrt{2}i}{2}\text{ and }x\neq \frac{\sqrt{2}i}{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$2fxx+f\times 1=1$$

Multiply x and x to get x^{2}.

$$2fx^{2}+f\times 1=1$$

Reorder the terms.

$$2fx^{2}+f=1$$

Combine all terms containing f.

$$\left(2x^{2}+1\right)f=1$$

Divide both sides by 2x^{2}+1.

$$\frac{\left(2x^{2}+1\right)f}{2x^{2}+1}=\frac{1}{2x^{2}+1}$$

Dividing by 2x^{2}+1 undoes the multiplication by 2x^{2}+1.

$$f=\frac{1}{2x^{2}+1}$$

Solve for f

$$f=\frac{1}{2x^{2}+1}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$2fxx+f\times 1=1$$

Multiply x and x to get x^{2}.

$$2fx^{2}+f\times 1=1$$

Reorder the terms.

$$2fx^{2}+f=1$$

Combine all terms containing f.

$$\left(2x^{2}+1\right)f=1$$

Divide both sides by 2x^{2}+1.

$$\frac{\left(2x^{2}+1\right)f}{2x^{2}+1}=\frac{1}{2x^{2}+1}$$

Dividing by 2x^{2}+1 undoes the multiplication by 2x^{2}+1.

$$f=\frac{1}{2x^{2}+1}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{-2+\frac{2}{f}}}{2}$$

$$x=\frac{\sqrt{-2+\frac{2}{f}}}{2}\text{, }f\neq 1\text{ and }f\neq 0$$

Solve for x

$$x=\frac{\sqrt{-2+\frac{2}{f}}}{2}$$

$$x=-\frac{\sqrt{-2+\frac{2}{f}}}{2}\text{, }f>0\text{ and }f<1$$