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

Problem 9 :

In holography, light from a laser beam is split into two beams, a reference beam and an object beam. Light from the object beam reflects off an object and is recombined with the reference beam to form images on film that can be used to create three-dimensional images.

a. Write an absolute value function that represents the path of the reference beam.

b. Write the function in part (a) as a piecewise function.

Solution :

The vertex of the path of the reference beam is (5, 8). So, the function has the form g(x) = a ∣ x − 5 ∣ + 8. Substitute the coordinates of the point (0, 0) into the equation and solve for a.

g(x) = a ∣ x − 5 ∣ + 8

When x = 0

0 = a ∣0 − 5∣ + 8

a = -1.6

So, the function g(x) = −1.6 ∣ x − 5 ∣ + 8 represents the path of the reference beam.

b. Write g(x) = −1.6 ∣ x − 5 ∣ + 8 as a piecewise function.

g(x) = -1.6[−(x − 5)] + 8, if x - 5 < 0

g(x) = -1.6(x − 5) + 8, if x - 5 ≥ 0

Simplify each expression and solve the inequalities. So, a piecewise function for 

g(x) = 1.6x    if x < 5

g(x) = -1.6x + 16    if  x ≥ 5

Problem 10 :

You are sitting on a boat on a lake. You can get a sunburn from the sunlight that hits you directly and also from the sunlight that reflects off the water.

a. Write an absolute value function that represents the path of the sunlight that reflects off the water.

b. Write the function in part (a) as a piecewise function.

Solution :

a) By observing the graph, it is clear that the vertex is at (3, 0)

y = a|x - h| + k

y = a|x - 3| + 0

y = a|x - 3|

Applying (2, 2), we get

2 = a|2 - 3|

2 = a(1)

a = 2

Applying the value of a, we get

y = 2|x - 3|

b) Defining as piece wise function.

f(x) = 2[-(x - 3)]  if x < 3

f(x) = 2[(x - 3)]  if x ≥ 3