Solve f^-1(x)=sqrt[3]{3+x}

$$f ^ { - 1 } ( x ) = \sqrt[ 3 ] { 3 + x }$$

Solve for f

$$f=\frac{x}{\sqrt[3]{x+3}}$$

$$x\neq 0\text{ and }x\neq -3$$

Steps for Solving Linear Equation

Reorder the terms.

$$\frac{1}{f}x=\sqrt[3]{x+3}$$

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

$$1x=f\sqrt[3]{x+3}$$

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

$$f\sqrt[3]{x+3}=1x$$

Reorder the terms.

$$\sqrt[3]{x+3}f=x$$

Divide both sides by \sqrt[3]{3+x}.

$$\frac{\sqrt[3]{x+3}f}{\sqrt[3]{x+3}}=\frac{x}{\sqrt[3]{x+3}}$$

Dividing by \sqrt[3]{3+x} undoes the multiplication by \sqrt[3]{3+x}.

$$f=\frac{x}{\sqrt[3]{x+3}}$$

Variable f cannot be equal to 0.

$$f=\frac{x}{\sqrt[3]{x+3}}\text{, }f\neq 0$$