Solve f^-1(x)=5-1x/3x

$$f ^ { - 1 } ( x ) = \frac { 5 - 1 x } { 3 x }$$

Solve for f

$$f=-\frac{3x^{2}}{x-5}$$

$$x\neq 0\text{ and }x\neq 5$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3x.

$$f^{-1}x\times 3x=5-x$$

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

$$f^{-1}x^{2}\times 3=5-x$$

Reorder the terms.

$$3\times \frac{1}{f}x^{2}=-x+5$$

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

$$3\times 1x^{2}=-xf+f\times 5$$

Multiply 3 and 1 to get 3.

$$3x^{2}=-xf+f\times 5$$

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

$$-xf+f\times 5=3x^{2}$$

Combine all terms containing f.

$$\left(-x+5\right)f=3x^{2}$$

The equation is in standard form.

$$\left(5-x\right)f=3x^{2}$$

Divide both sides by -x+5.

$$\frac{\left(5-x\right)f}{5-x}=\frac{3x^{2}}{5-x}$$

Dividing by -x+5 undoes the multiplication by -x+5.

$$f=\frac{3x^{2}}{5-x}$$

Variable f cannot be equal to 0.

$$f=\frac{3x^{2}}{5-x}\text{, }f\neq 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{f\left(f+60\right)}-f}{6}$$

$$x=\frac{-\sqrt{f\left(f+60\right)}-f}{6}\text{, }f\neq 0$$

Solve for x

$$x=\frac{\sqrt{f\left(f+60\right)}-f}{6}$$

$$x=\frac{-\sqrt{f\left(f+60\right)}-f}{6}\text{, }f>0\text{ or }f\leq -60$$