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

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

Solve for f

$$f=\frac{x}{\sqrt{x+1}}$$

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

Steps for Solving Linear Equation

Reorder the terms.

$$\frac{1}{f}x=\sqrt{x+1}$$

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

$$1x=f\sqrt{x+1}$$

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

$$f\sqrt{x+1}=1x$$

Reorder the terms.

$$\sqrt{x+1}f=x$$

Divide both sides by \sqrt{x+1}.

$$\frac{\sqrt{x+1}f}{\sqrt{x+1}}=\frac{x}{\sqrt{x+1}}$$

Dividing by \sqrt{x+1} undoes the multiplication by \sqrt{x+1}.

$$f=\frac{x}{\sqrt{x+1}}$$

Variable f cannot be equal to 0.

$$f=\frac{x}{\sqrt{x+1}}\text{, }f\neq 0$$

Solve for x

$$\left\{\begin{matrix}x=\frac{f\left(\sqrt{f^{2}+4}+f\right)}{2}\text{, }&\frac{f\sqrt{f^{2}+4}}{2}+\frac{f^{2}}{2}+1\geq 0\text{ and }f\neq 0\text{ and }\frac{f\sqrt{f^{2}+4}+f^{2}}{2}\geq -1\\x=\frac{f\left(-\sqrt{f^{2}+4}+f\right)}{2}\text{, }&-\frac{f\sqrt{f^{2}+4}}{2}+\frac{f^{2}}{2}+1\geq 0\text{ and }f\left(-\sqrt{f^{2}+4}+f\right)<-2\text{ and }f\neq 0\text{ and }\frac{-f\sqrt{f^{2}+4}+f^{2}}{2}\geq -1\end{matrix}\right.$$