Find the Inverse x=2|y|-1

$x = 2 | y | - 1$

Step 1

Rewrite the equation as $2 | y | - 1 = x$ .

$2 | y | - 1 = x$

Step 2

Add $1$ to both sides of the equation.

$2 | y | = x + 1$

Step 3

Divide each term in $2 | y | = x + 1$ by $2$ and simplify.

Divide each term in $2 | y | = x + 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $| y |$ by $1$ .

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

Step 4

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

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

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Next, use the negative value of the $\pm$ to find the second solution.

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

The complete solution is the result of both the positive and negative portions of the solution.

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

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

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

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

Step 6

Apply the distributive property.

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

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

Step 7

Interchange the variables. Create an equation for each expression.

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

Step 8

Solve each equation for $y$ .

Solve for $y$ .

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

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

Subtract $\frac{1}{2}$ from both sides of the equation.

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$y = 2 x - 1$

Solve for $y$ .

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

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

Add $\frac{1}{2}$ to both sides of the equation.

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

Multiply both sides of the equation by $- 2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 2 (- \frac{y}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{y}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $y$ by $1$ .

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

Simplify the right side.

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y = - 2 x - 1$

List the equations.

$y = 2 x - 1, y = - 2 x - 1$

Step 9

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

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

Step 10

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

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{x}{2} + \frac{1}{2}, - \frac{x}{2} - \frac{1}{2}$ and $f^{- 1} (x) = 2 x - 1, f^{- 1} (x) = - 2 x - 1$ and compare them.

Find the range of $\frac{x}{2} + \frac{1}{2}$ .

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \infty)$

Find the domain of $2 x - 1$ .

$(- \infty, \infty)$

Find the domain of $\frac{x}{2} + \frac{1}{2}$ .

$(- \infty, \infty)$

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

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

Step 11

Find the Inverse x=2|y|-1

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