writing absolute value as piecewise - taisee island
Post on 22-Mar-2022
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Writing Absolute Value as Piecewise Piecewise: A graph split into multiple sections or written as multiple parts.
f(x) = |x|
π π₯ = π₯, π₯ β€ 0βπ₯, π₯ β₯ 0
Write f(x) = |x β 1| + 2 as a piecewise function.
First find the split. It occurs where the argument equals zero. x β 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) β 1 -1 This is negative
Right of 1: Let x = 2 (2) β 1 1 This is positive
+ β 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 β 1 + 2
-x + 3 x + 1
The simplified negative equation is for x β€ 1 while the positive equation is for x β₯ 1 according to the sign chart.
π π₯ = βπ₯ + 3, π₯ β€ 1π₯ + 1, π₯ β₯ 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
+ β 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 β 9 + 5 3x + 14
3x β 4
The simplified negative equation is for x β€ -3 while the positive equation is for x β₯ -3 according to the sign chart.
π π₯ = 3π₯ β 4, π₯ β€ β33π₯ + 14, π₯ β₯ β3
Write f(x) = 5|3 β x| as a piecewise function.
First find the split. It occurs where the argument equals zero.
3 β 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 β (0) 3 This is positive
Right of 3: Let x = 5 3 β 5 -2 This is negative
β + 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 β x)
5x β 15 -5x + 15
The simplified negative equation is for x β₯ 3 while the positive equation is for x β€ 3 according to the sign chart.
π π₯ = β5π₯ + 15, π₯ β€ 35π₯ β 15, π₯ β₯ 3
Write f(x) = |2x2 + 5x β 3| as a piecewise function.
First find the split(s). They occur when the argument equals zero.
2x2 + 5x β 3 = 0
(2x β 1)(x + 3) = 0
2x β 1 = 0 and x + 3 = 0
x = Β½ and x = -3
2(-4)2 + 5(-4) β 3 2(0)2 + 5(0) β 3 2(1)2 + 5(1) β 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 Β½, and greater than Β½.
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