Writing an Absolute Value Piecewise

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Idea

Absolute value measures distance from zero, so its result is nonnegative. A negative expression must have its sign reversed, while a nonnegative expression is left alone.

Method

|u|=\begin{cases}u,&u\ge0,\\-u,&u<0.\end{cases}

For an expression $u(x)$ , solve $u(x)=0$ to locate the breakpoints before splitting into intervals.

Worked example

For $|2x-6|$ , the breakpoint is $x=3$ . If $x<3$ , then $2x-6<0$ and the value is $6-2x$ . If $x\ge3$ , the value is $2x-6$ . Both branches meet at zero when $x=3$ , producing a corner rather than a discontinuity.

Common mistake

$|a+b|$ is not generally $|a|+|b|$ . For $a=1,b=-1$ , the two sides are $0$ and $2$ .

Check your understanding

Write $|x+1|$ piecewise.

Show answer

$-x-1$ for $x<-1$ , and $x+1$ for $x\ge-1$ .

Explore

Domain and range

Try this. Switch functions and compare allowed inputs and outputs. Open circles exclude endpoints; filled circles include them.

Function

y = |x| · Domain: all real x · Range: y ≥ 0

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

When x < 3, which expression equals |x−3|?

x−3

3+x

3−x

Hint 1 · Find a starting point

The expression inside the absolute value is negative.

Hint 2 · Take the next step

For a negative a, |a|=−a.

Show the reasoning

Answer: 3−x

−(x−3)=3−x on this interval.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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