Precalculus Examples

$f (x) = | x - 2 | + 3$

Step 1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x - 2 \geq 0$

Step 2

Add $2$ to both sides of the inequality.

$x \geq 2$

Step 3

In the piece where $x - 2$ is non-negative, remove the absolute value.

$x - 2 + 3$

Step 4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x - 2 < 0$

Step 5

Add $2$ to both sides of the inequality.

$x < 2$

Step 6

In the piece where $x - 2$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x - 2) + 3$

Step 7

Write as a piecewise.

$f (x) = {\begin{cases} x - 2 + 3 & x \geq 2 \\ - (x - 2) + 3 & x < 2 \end{cases}$

Step 8

Add $- 2$ and $3$ .

$f (x) = {\begin{cases} x + 1 & x \geq 2 \\ - (x - 2) + 3 & x < 2 \end{cases}$

Step 9

Simplify $- (x - 2) + 3$ .

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Step 9.1

Simplify each term.

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Step 9.1.1

Apply the distributive property.

$f (x) = {\begin{cases} x + 1 & x \geq 2 \\ - x + 2 + 3 & x < 2 \end{cases}$

Step 9.1.2

Multiply $- 1$ by $- 2$ .

$f (x) = {\begin{cases} x + 1 & x \geq 2 \\ - x + 2 + 3 & x < 2 \end{cases}$

Step 9.2

Add $2$ and $3$ .

$f (x) = {\begin{cases} x + 1 & x \geq 2 \\ - x + 5 & x < 2 \end{cases}$