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

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

Solve for f

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

$$x\neq 0\text{ and }x\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x.

$$f^{-1}x\times 2x=1-x$$

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

$$f^{-1}x^{2}\times 2=1-x$$

Reorder the terms.

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

Variable f cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by f.

$$2\times 1x^{2}=-xf+f$$

Multiply 2 and 1 to get 2.

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

Swap sides so that all variable terms are on the left hand side.

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

Combine all terms containing f.

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

The equation is in standard form.

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

Divide both sides by -x+1.

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

Dividing by -x+1 undoes the multiplication by -x+1.

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

Variable f cannot be equal to 0.

$$f=\frac{2x^{2}}{1-x}\text{, }f\neq 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{f\left(f+8\right)}-f}{4}$$

$$x=\frac{-\sqrt{f\left(f+8\right)}-f}{4}\text{, }f\neq 0$$

Solve for x

$$x=\frac{\sqrt{f\left(f+8\right)}-f}{4}$$

$$x=\frac{-\sqrt{f\left(f+8\right)}-f}{4}\text{, }f>0\text{ or }f\leq -8$$