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

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

Solve for f

$$f=x^{2}-\frac{x}{2}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$2f\times 1=2xx+x\left(-1\right)$$

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

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

Multiply 2 and 1 to get 2.

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

The equation is in standard form.

$$2f=2x^{2}-x$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide x\left(-1+2x\right) by 2.

$$f=x^{2}-\frac{x}{2}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{\sqrt{16f+1}+1}{4}\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{16f+1}+1}{4}\text{, }&f\neq 0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{16f+1}+1}{4}\text{, }&f\geq -\frac{1}{16}\\x=\frac{-\sqrt{16f+1}+1}{4}\text{, }&f\neq 0\text{ and }f\geq -\frac{1}{16}\end{matrix}\right.$$