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

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

Solve for f

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

$$x\neq 2\text{ and }x\neq 0\text{ and }x\neq -1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x-2.

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

Use the distributive property to multiply f^{-1}x by x-2.

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

Reorder the terms.

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

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

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

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

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

Reorder the terms.

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

Combine all terms containing f.

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

Divide both sides by -x-1.

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

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

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

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

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

Variable f cannot be equal to 0.

$$f=-\frac{x\left(x-2\right)}{x+1}\text{, }f\neq 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{f^{2}-8f+4}}{2}-\frac{f}{2}+1$$

$$x=-\frac{\sqrt{f^{2}-8f+4}}{2}-\frac{f}{2}+1\text{, }f\neq 0$$

Solve for x

$$x=\frac{\sqrt{f^{2}-8f+4}}{2}-\frac{f}{2}+1$$

$$x=-\frac{\sqrt{f^{2}-8f+4}}{2}-\frac{f}{2}+1\text{, }\left(f\neq 0\text{ and }f\leq 4-2\sqrt{3}\right)\text{ or }f\geq 2\sqrt{3}+4$$