writing absolute value as piecewise - taisee island
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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.