Solve f^-1(x)=2x^2+1/sqrt{x}

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

Solve for f

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

$$x>0$$

Steps for Solving Linear Equation

Reorder the terms.

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

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

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

Use the distributive property to multiply fx^{-\frac{1}{2}} by 2x^{2}+1.

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

To multiply powers of the same base, add their exponents. Add -\frac{1}{2} and 2 to get \frac{3}{2}.

$$1x=2fx^{\frac{3}{2}}+fx^{-\frac{1}{2}}$$

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

$$2fx^{\frac{3}{2}}+fx^{-\frac{1}{2}}=1x$$

Reorder the terms.

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

Combine all terms containing f.

$$\left(2x^{\frac{3}{2}}+x^{-\frac{1}{2}}\right)f=x$$

The equation is in standard form.

$$\left(2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}\right)f=x$$

Divide both sides by 2x^{\frac{3}{2}}+x^{-\frac{1}{2}}.

$$\frac{\left(2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}\right)f}{2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}}=\frac{x}{2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}}$$

Dividing by 2x^{\frac{3}{2}}+x^{-\frac{1}{2}} undoes the multiplication by 2x^{\frac{3}{2}}+x^{-\frac{1}{2}}.

$$f=\frac{x}{2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}}$$

Divide x by 2x^{\frac{3}{2}}+x^{-\frac{1}{2}}.

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

Variable f cannot be equal to 0.

$$f=\frac{x^{\frac{3}{2}}}{2x^{2}+1}\text{, }f\neq 0$$