Trigonometry Examples

$y = | x - 1 | + 2$

Step 1

Interchange the variables.

$x = | y - 1 | + 2$

Step 2

Solve for $y$ .

Tap for more steps...

Step 2.1

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

$| y - 1 | + 2 = x$

Step 2.2

Subtract $2$ from both sides of the equation.

$| y - 1 | = x - 2$

Step 2.3

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

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

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$y - 1 = x - 2$

Step 2.4.2

Move all terms not containing $y$ to the right side of the equation.

Tap for more steps...

Step 2.4.2.1

Add $1$ to both sides of the equation.

$y = x - 2 + 1$

Step 2.4.2.2

Add $- 2$ and $1$ .

$y = x - 1$

Step 2.4.3

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

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

Step 2.4.4

Simplify $- (x - 2)$ .

Tap for more steps...

Step 2.4.4.1

Rewrite.

$y - 1 = 0 + 0 - (x - 2)$

Step 2.4.4.2

Simplify by adding zeros.

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

Step 2.4.4.3

Apply the distributive property.

$y - 1 = - x - - 2$

Step 2.4.4.4

Multiply $- 1$ by $- 2$ .

$y - 1 = - x + 2$

Step 2.4.5

Move all terms not containing $y$ to the right side of the equation.

Tap for more steps...

Step 2.4.5.1

Add $1$ to both sides of the equation.

$y = - x + 2 + 1$

Step 2.4.5.2

Add $2$ and $1$ .

$y = - x + 3$

Step 2.4.6

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

$y = x - 1$

$y = - x + 3$

$y = x - 1$

$y = - x + 3$

$y = x - 1$

$y = - x + 3$

Step 3

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

$f^{- 1} (x) = x - 1, - x + 3$

Step 4

Verify if $f^{- 1} (x) = x - 1, - x + 3$ is the inverse of $f (x) = | x - 1 | + 2$ .

Tap for more steps...

Step 4.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = | x - 1 | + 2$ and $f^{- 1} (x) = x - 1, - x + 3$ and compare them.

Step 4.2

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

Tap for more steps...

Step 4.2.1

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

Interval Notation:

$[2, \infty)$

Step 4.3

Find the domain of $x - 1$ .

Tap for more steps...

Step 4.3.1

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 4.4

Since the domain of $f^{- 1} (x) = x - 1, - x + 3$ is not equal to the range of $f (x) = | x - 1 | + 2$ , then $f^{- 1} (x) = x - 1, - x + 3$ is not an inverse of $f (x) = | x - 1 | + 2$ .

There is no inverse

Step 5