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

$f (x) = | 2 x - 1 |$

Write $f (x) = | 2 x - 1 |$ as an equation.

$y = | 2 x - 1 |$

Interchange the variables.

$x = | 2 y - 1 |$

Solve for $y$ .

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

$| 2 y - 1 | = x$

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

$2 y - 1 = \pm x$

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.

$2 y - 1 = x$

Add $1$ to both sides of the equation.

$2 y = x + 1$

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 $1$ .

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

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

$2 y - 1 = - x$

Add $1$ to both sides of the equation.

$2 y = - x + 1$

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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

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

Find the range of $f (x) = | 2 x - 1 |$ .

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

Interval Notation:

$[0, \infty)$

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

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Since the domain of $f^{- 1} (x) = \frac{x}{2} + \frac{1}{2}, - \frac{x}{2} + \frac{1}{2}$ is not equal to the range of $f (x) = | 2 x - 1 |$ , then $f^{- 1} (x) = \frac{x}{2} + \frac{1}{2}, - \frac{x}{2} + \frac{1}{2}$ is not an inverse of $f (x) = | 2 x - 1 |$ .

There is no inverse

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

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