Find the Inverse y=(1- square root of x)/(1+ square root of x)

$y = \frac{1 - \sqrt{x}}{1 + \sqrt{x}}$

Interchange the variables.

$x = \frac{1 - \sqrt{y}}{1 + \sqrt{y}}$

Solve for $y$ .

Rewrite the equation as $\frac{1 - \sqrt{y}}{1 + \sqrt{y}} = x$ .

$\frac{1 - \sqrt{y}}{1 + \sqrt{y}} = x$

Multiply both sides by $1 + \sqrt{y}$ .

$\frac{1 - \sqrt{y}}{1 + \sqrt{y}} (1 + \sqrt{y}) = x (1 + \sqrt{y})$

Simplify.

Simplify the left side.

Cancel the common factor of $1 + \sqrt{y}$ .

Cancel the common factor.

$\frac{1 - \sqrt{y}}{1 + \sqrt{y}} (1 + \sqrt{y}) = x (1 + \sqrt{y})$

Rewrite the expression.

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

Simplify the right side.

Simplify $x (1 + \sqrt{y})$ .

Apply the distributive property.

$1 - \sqrt{y} = x \cdot 1 + x \sqrt{y}$

Multiply $x$ by $1$ .

$1 - \sqrt{y} = x + x \sqrt{y}$

Solve for $y$ .

Solve for $\sqrt{y}$ .

Subtract $x \sqrt{y}$ from both sides of the equation.

$1 - \sqrt{y} - x \sqrt{y} = x$

Subtract $1$ from both sides of the equation.

$- \sqrt{y} - x \sqrt{y} = x - 1$

Factor $\sqrt{y}$ out of $- \sqrt{y} - x \sqrt{y}$ .

Factor $\sqrt{y}$ out of $- \sqrt{y}$ .

$\sqrt{y} \cdot - 1 - x \sqrt{y} = x - 1$

Factor $\sqrt{y}$ out of $- x \sqrt{y}$ .

$\sqrt{y} \cdot - 1 + \sqrt{y} (- x) = x - 1$

Factor $\sqrt{y}$ out of $\sqrt{y} \cdot - 1 + \sqrt{y} (- x)$ .

$\sqrt{y} (- 1 - x) = x - 1$

Divide each term in $\sqrt{y} (- 1 - x) = x - 1$ by $- 1 - x$ and simplify.

Divide each term in $\sqrt{y} (- 1 - x) = x - 1$ by $- 1 - x$ .

$\frac{\sqrt{y} (- 1 - x)}{- 1 - x} = \frac{x}{- 1 - x} + \frac{- 1}{- 1 - x}$

Simplify the left side.

Cancel the common factor of $- 1 - x$ .

Cancel the common factor.

$\frac{\sqrt{y} (- 1 - x)}{- 1 - x} = \frac{x}{- 1 - x} + \frac{- 1}{- 1 - x}$

Divide $\sqrt{y}$ by $1$ .

$\sqrt{y} = \frac{x}{- 1 - x} + \frac{- 1}{- 1 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$\sqrt{y} = \frac{x - 1}{- 1 - x}$

Rewrite $- 1$ as $- 1 (1)$ .

$\sqrt{y} = \frac{x - 1}{- 1 (1) - x}$

Factor $- 1$ out of $- x$ .

$\sqrt{y} = \frac{x - 1}{- 1 (1) - (x)}$

Factor $- 1$ out of $- 1 (1) - (x)$ .

$\sqrt{y} = \frac{x - 1}{- 1 (1 + x)}$

Move the negative in front of the fraction.

$\sqrt{y} = - \frac{x - 1}{1 + x}$

To remove the radical on the left side of the equation, square both sides of the equation.

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y}$ as $y^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${(y^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(y^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y^{\frac{1}{2} \cdot 2} = {(- \frac{x - 1}{1 + x})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y^{\frac{1}{2} \cdot 2} = {(- \frac{x - 1}{1 + x})}^{2}$

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(- \frac{x - 1}{1 + x})}^{2}$ .

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{x - 1}{1 + x}$ .

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

Apply the product rule to $\frac{x - 1}{1 + x}$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply $\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}$ by $1$ .

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

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

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

Verify if $f^{- 1} (x) = \frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}$ is the inverse of $f (x) = \frac{1 - \sqrt{x}}{1 + \sqrt{x}}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Remove parentheses.

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

Simplify the numerator.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{1 + \sqrt{x}}{1 + \sqrt{x}}$ .

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

Combine $- 1$ and $\frac{1 + \sqrt{x}}{1 + \sqrt{x}}$ .

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{1 - \sqrt{x} - (1 + \sqrt{x})}{1 + \sqrt{x}}$ in a factored form.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Let $u = x^{\frac{1}{2}}$ . Substitute $u$ for all occurrences of $x^{\frac{1}{2}}$ .

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

Replace all occurrences of $u$ with $x^{\frac{1}{2}}$ .

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

Apply the product rule to $\frac{- 2 x^{\frac{1}{2}}}{1 + \sqrt{x}}$ .

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

Simplify the numerator.

Apply the product rule to $- 2 x^{\frac{1}{2}}$ .

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

Raise $- 2$ to the power of $2$ .

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

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x^{\frac{1}{2} \cdot 2}}{{(1 + \sqrt{x})}^{2}}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x^{\frac{1}{2} \cdot 2}}{{(1 + \sqrt{x})}^{2}}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Rewrite the expression.

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

Simplify.

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

Rewrite ${(1 + \sqrt{x})}^{2}$ as $(1 + \sqrt{x}) (1 + \sqrt{x})$ .

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

Expand $(1 + \sqrt{x}) (1 + \sqrt{x})$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 \cdot 1 + 1 \sqrt{x} + \sqrt{x} (1 + \sqrt{x})}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Apply the distributive property.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 \cdot 1 + 1 \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \sqrt{x}}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 1 \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \sqrt{x}}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Multiply $\sqrt{x}$ by $1$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \sqrt{x}}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Multiply $\sqrt{x}$ by $1$ .

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

Multiply $\sqrt{x} \sqrt{x}$ .

Raise $\sqrt{x}$ to the power of $1$ .

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

Raise $\sqrt{x}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + \sqrt{x} + \sqrt{x} + x^{\frac{1}{2} \cdot 2}}}{{(1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}$

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Add $\sqrt{x}$ and $\sqrt{x}$ .

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

Simplify the denominator.

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{1 + \sqrt{x} + 1 - \sqrt{x}}{1 + \sqrt{x}}$ in a factored form.

Add $1$ and $1$ .

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

Subtract $\sqrt{x}$ from $\sqrt{x}$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 2 \sqrt{x} + x}}{{(\frac{2 + 0}{1 + \sqrt{x}})}^{2}}$

Add $2$ and $0$ .

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

Apply the product rule to $\frac{2}{1 + \sqrt{x}}$ .

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

Raise $2$ to the power of $2$ .

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

Rewrite ${(1 + \sqrt{x})}^{2}$ as $(1 + \sqrt{x}) (1 + \sqrt{x})$ .

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

Expand $(1 + \sqrt{x}) (1 + \sqrt{x})$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 2 \sqrt{x} + x}}{\frac{4}{1 \cdot 1 + 1 \sqrt{x} + \sqrt{x} (1 + \sqrt{x})}}$

Apply the distributive property.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 2 \sqrt{x} + x}}{\frac{4}{1 \cdot 1 + 1 \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \sqrt{x}}}$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 2 \sqrt{x} + x}}{\frac{4}{1 + 1 \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \sqrt{x}}}$

Multiply $\sqrt{x}$ by $1$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 2 \sqrt{x} + x}}{\frac{4}{1 + \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \sqrt{x}}}$

Multiply $\sqrt{x}$ by $1$ .

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

Multiply $\sqrt{x} \sqrt{x}$ .

Raise $\sqrt{x}$ to the power of $1$ .

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

Raise $\sqrt{x}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{\frac{4 x}{1 + 2 \sqrt{x} + x}}{\frac{4}{1 + \sqrt{x} + \sqrt{x} + x^{\frac{1}{2} \cdot 2}}}$

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Add $\sqrt{x}$ and $\sqrt{x}$ .

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

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{4 x}{1 + 2 \sqrt{x} + x} \cdot \frac{1 + 2 \sqrt{x} + x}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{4 (x)}{1 + 2 \sqrt{x} + x} \cdot \frac{1 + 2 \sqrt{x} + x}{4}$

Cancel the common factor.

$f^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = \frac{4 x}{1 + 2 \sqrt{x} + x} \cdot \frac{1 + 2 \sqrt{x} + x}{4}$

Rewrite the expression.

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

Cancel the common factor of $1 + 2 \sqrt{x} + x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify the numerator.

Rewrite $\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}$ as ${(\frac{x - 1}{1 + x})}^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x + 1 - (x - 1)}{x + 1}$ in a factored form.

Apply the distributive property.

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $x$ .

$f (\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}) = \frac{\frac{0 + 1 + 1}{x + 1}}{1 + \sqrt{\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}}}$

Add $0$ and $1$ .

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

Add $1$ and $1$ .

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

Simplify the denominator.

Rewrite $\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}$ as ${(\frac{x - 1}{1 + x})}^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x + x + 1 - 1}{x + 1}$ in a factored form.

Add $x$ and $x$ .

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

Subtract $1$ from $1$ .

$f (\frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}) = \frac{\frac{2}{x + 1}}{\frac{2 x + 0}{x + 1}}$

Add $2 x$ and $0$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{{(x - 1)}^{2}}{{(1 + x)}^{2}}$ is the inverse of $f (x) = \frac{1 - \sqrt{x}}{1 + \sqrt{x}}$ .

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

Find the Inverse y=(1- square root of x)/(1+ square root of x)

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