Find the Inverse f(x)=(1-x)/(2x)

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

Step 1

Write $f (x) = \frac{1 - x}{2 x}$ as an equation.

$y = \frac{1 - x}{2 x}$

Step 2

Interchange the variables.

$x = \frac{1 - y}{2 y}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1 - y}{2 y} = x$ .

$\frac{1 - y}{2 y} = x$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 y, 1$

The LCM of one and any expression is the expression.

$2 y$

Multiply each term in $\frac{1 - y}{2 y} = x$ by $2 y$ to eliminate the fractions.

Multiply each term in $\frac{1 - y}{2 y} = x$ by $2 y$ .

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

Simplify the left side.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Rewrite the expression.

$1 - y = x (2 y)$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$1 - y = 2 x y$

Solve the equation.

Subtract $2 x y$ from both sides of the equation.

$1 - y - 2 x y = 0$

Subtract $1$ from both sides of the equation.

$- y - 2 x y = - 1$

Factor $y$ out of $- y - 2 x y$ .

Factor $y$ out of $- y$ .

$y \cdot - 1 - 2 x y = - 1$

Factor $y$ out of $- 2 x y$ .

$y \cdot - 1 + y (- 2 x) = - 1$

Factor $y$ out of $y \cdot - 1 + y (- 2 x)$ .

$y (- 1 - 2 x) = - 1$

Divide each term in $y (- 1 - 2 x) = - 1$ by $- 1 - 2 x$ and simplify.

Divide each term in $y (- 1 - 2 x) = - 1$ by $- 1 - 2 x$ .

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

Simplify the left side.

Cancel the common factor of $- 1 - 2 x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{- 1}{- 1 - 2 x}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{1}{- 1 - 2 x}$

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

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

Factor $- 1$ out of $- 2 x$ .

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

Factor $- 1$ out of $- 1 (1) - (2 x)$ .

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

Simplify the expression.

Move the negative in front of the fraction.

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

Multiply $- 1$ by $- 1$ .

$y = 1 \frac{1}{1 + 2 x}$

Multiply $\frac{1}{1 + 2 x}$ by $1$ .

$y = \frac{1}{1 + 2 x}$

Step 4

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

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

Step 5

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

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

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

Simplify the denominator.

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{2 x + 1 - 1}{2 x + 1}$ in a factored form.

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{x}{2 x + 1}$ by $1 + 2 x$ .

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

Cancel the common factor of $1 + 2 x$ and $2 x + 1$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse f(x)=(1-x)/(2x)

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