Precalculus Examples

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

Step 1

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

$y = | x - 2 |$

Step 2

Interchange the variables.

$x = | y - 2 |$

Step 3

Solve for $y$ .

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Step 3.1

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

$| y - 2 | = x$

Step 3.2

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

$y - 2 = \pm x$

Step 3.3

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

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Step 3.3.1

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

$y - 2 = x$

Step 3.3.2

Add $2$ to both sides of the equation.

$y = x + 2$

Step 3.3.3

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

$y - 2 = - x$

Step 3.3.4

Add $2$ to both sides of the equation.

$y = - x + 2$

Step 3.3.5

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

$y = x + 2$

$y = - x + 2$

$y = x + 2$

$y = - x + 2$

$y = x + 2$

$y = - x + 2$

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

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

Interval Notation:

$[0, \infty)$

Step 5.3

Find the domain of $x + 2$ .

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Step 5.3.1

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.

$(- \infty, \infty)$

Step 5.4

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

There is no inverse

Step 6