# CONVERTING ABSOLUTE VALUE FUNCTIONS TO PIECEWISE DEFINED FUNCTION

For each of the functions, create a piecewise definition.

Problem 1 :

f(x) = |x - 1|

Solution :

f(x) = |x - 1|

f(x) = x - 1  and  f(x) = -(x - 1)

Case 1 :

f(x) = x - 1

when x ≥ 1, f(x) is positive

Case 2 :

f(x) = -(x - 1)

when x < 1, f(x) is negative

Distributing negative, we get

f(x) = - x + 1

So, the required definition is

Problem 2 :

f(x) = |x + 2|

Solution :

f(x) = |x + 2|

f(x) = x + 2  and  f(x) = -(x + 2)

Let f(x) = 0

x + 2 = 0

x = -2

Case 1 :

f(x) = x + 2

when x ≥ -2, f(x) is positive

Case 2 :

f(x) = -(x + 2)

when x < -2, f(x) is negative

Distributing negative, we get

f(x) = - x - 2

Problem 3 :

f(x) = |2x - 1|

Solution :

f(x) = |2x - 1|

f(x) = 2x - 1  and  f(x) = -(2x - 1)

Let f(x) = 0

2x - 1 = 0

x = 1/2

Case 1 :

f(x) = 2x - 1

when x ≥ 1/2, f(x) is positive

Case 2 :

f(x) = -(2x - 1)

when x < 1/2, f(x) is negative

Distributing negative, we get

f(x) = - 2x + 1

Problem 4 :

f(x) = |5 - 2x|

Solution :

f(x) = |5 - 2x|

f(x) = 5 - 2x  and  f(x) = -(5 - 2x)

Let f(x) = 0

5 - 2x = 0

2x = 5

x = 5/2

Case 1 :

f(x) = 5 - 2x

when x  5/2, f(x) is positive

Case 2 :

f(x) = -(5 - 2x)

when x > 5/2, f(x) is negative

Distributing negative, we get

f(x) = -5 + 2x or 2x - 5

Problem 5 :

f(x) = |1 - 3x|

Solution :

f(x) = |1 - 3x|

f(x) = 1 - 3x  and  f(x) = -(1 - 3x)

Let f(x) = 0

1 - 3x = 0

3x = 1

x = 1/3

Case 1 :

f(x) = 1 - 3x

when x ≤ 1/3, f(x) is positive

Case 2 :

f(x) = -(1 - 3x)

when x > 1/3, f(x) is negative

Distributing negative, we get

f(x) = -1 + 3x or 3x - 1

Problem 6 :

f(x) = |2x + 1|

Solution :

f(x) = |2x + 1|

f(x) = 2x + 1  and  f(x) = -(2x + 1)

Let f(x) = 0

2x + 1 = 0

2x = -1

x = -1/2

Case 1 :

f(x) = 2x + 1

when x ≥ -1/2, f(x) is positive

Case 2 :

f(x) = -(2x + 1)

when x < -1/2, f(x) is negative

Distributing negative, we get

f(x) = -2x - 1

Problem 7 :

f(x) = x - |x|

Solution :

f(x) = x - |x|

Case 1 :

f(x) = x - x

f(x) = 0

Case 2 :

f(x) = x - (-x)

= x + x

f(x) = 2x

Problem 8 :

f(x) = x + |x - 1|

Solution :

f(x) = x + |x - 1|

Case 1 :

f(x) = x + x - 1

f(x) = 2x - 1

Case 2 :

f(x) = x + (-x + 1)

= x - x + 1

f(x) = 1

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