Writing Absolute Value as Piecewise Piecewise: A graph split into multiple sections or written as multiple parts.

f(x) = |x|

𝑓 𝑥 = 𝑥, 𝑥 ≤ 0−𝑥, 𝑥 ≥ 0

Write f(x) = |x – 1| + 2 as a piecewise function.

First find the split. It occurs where the argument equals zero. x – 1 = 0 when x = 1. The split is at x=1.

Determine where the graph is positive vs. negative (reflected).

Substitute values into the argument to see if the outcome is positive or negative.

Chose a value to the left of 1 and a value to the right of 1.

1

Left of 1: Let x = 0 (0) – 1 -1 This is negative

Right of 1: Let x = 2 (2) – 1 1 This is positive

+ – Sign Chart

For the positive side, drop absolute value and simplify.

For the negative side, switch the signs of the argument & simplify.

Negative Positive

(-x + 1) + 2 x – 1 + 2

-x + 3 x + 1

The simplified negative equation is for x ≤ 1 while the positive equation is for x ≥ 1 according to the sign chart.

𝑓 𝑥 = −𝑥 + 3, 𝑥 ≤ 1𝑥 + 1, 𝑥 ≥ 1

Write f(x) = -|3x + 9| + 5 as a piecewise function.

First find the split. It occurs where the argument equals zero. 3x + 9 = 0 when x = -3.

The split is at x = -3.

Determine where the graph is positive vs. negative (reflected).

Substitute values into the argument to see if the outcome is positive or negative.

Chose a value to the left of -3 and a value to the right of -3.

-3

Left of -3: Let x = -4 3(-4) + 9 -3 This is negative

Right of -3: Let x = 0 3(0) + 9 9 This is positive

+ – Sign Chart

For the positive side, drop absolute value and simplify.

For the negative side, switch the signs of the argument & simplify.

Negative Positive

-(-3x + 9) + 5 3x + 9 + 5

3x – 9 + 5 3x + 14

3x – 4

The simplified negative equation is for x ≤ -3 while the positive equation is for x ≥ -3 according to the sign chart.

𝑓 𝑥 = 3𝑥 − 4, 𝑥 ≤ −33𝑥 + 14, 𝑥 ≥ −3

Write f(x) = 5|3 – x| as a piecewise function.

First find the split. It occurs where the argument equals zero.

3 – x = 0

x = 3

The split is at x = 3.

Determine where the graph is positive vs. negative (reflected).

Substitute values into the argument to see if the outcome is positive or negative.

Chose a value to the left of 3 and a value to the right of 3.

3

Left of 3: Let x = 0 3 – (0) 3 This is positive

Right of 3: Let x = 5 3 – 5 -2 This is negative

– + Sign Chart

For the positive side, drop absolute value and simplify.

For the negative side, switch the signs of the argument & simplify.

Negative Positive

5(-3 + x) 5(3 – x)

5x – 15 -5x + 15

The simplified negative equation is for x ≥ 3 while the positive equation is for x ≤ 3 according to the sign chart.

𝑓 𝑥 = −5𝑥 + 15, 𝑥 ≤ 35𝑥 − 15, 𝑥 ≥ 3

Write f(x) = |2x2 + 5x – 3| as a piecewise function.

First find the split(s). They occur when the argument equals zero.

2x2 + 5x – 3 = 0

(2x – 1)(x + 3) = 0

2x – 1 = 0 and x + 3 = 0

x = ½ and x = -3

2(-4)2 + 5(-4) – 3 2(0)2 + 5(0) – 3 2(1)2 + 5(1) – 3

9 (positive) -3 (negative) 4 (positive)

Only the middle section needs to be reflected (put a negative in front).

Determine where the graph is positive vs. negative (reflected).

Check the intervals: less than -3, in between -3 and ½, and greater than ½.

Simplify.

2x2 + 5x – 3 -(2x2 + 5x – 3)

-2x2 – 5x + 3

Alternatively, f(x) can be written:

2

2

2

2 5 3, 3

( ) 2 5 3, 3 x 1/ 2

2 5 3, 1/ 2

x x x

f x x x

x x x

2

2

2 5 3, 3 and 1/ 2( )

2 5 3, 3 x 1/ 2

x x x xf x

x x