WRITING ABSOLUTE VALUE FUNCTIONS AS PIECEWISE FUNCTIONS WORKSHEET

Write each of the following absolute value functions as a piecewise function.

Problem 1 :

f(x) = |x|

Problem 2 :

f(x) = |2x - 3|

Problem 3 :

f(x) = |3 - x| + 5

Problem 4 :

f(x) = |x + 2| + |x - 3|

Problem 5 :

f(x) = |x - 5| + |x|

Problem 6 :

Problem 7 :

Problem 8 :

Answers

1. Answer :

f(x) = |x|

Equate the expression inside the absolute value to zero and solve for x.

x = 0

When x takes a negative value, x < 0.

f(x) = -x

When x takes zero or a positive value, x ≥ 0.

f(x) = x

Therefore,

2. Answer :

f(x) = |2x - 3|

Equate the expression inside the absolute value to zero and solve for x.

2x - 3 = 0

2x = 3

x = 1.5

When x < 1.5, (2x - 3) < 0.

f(x) = -(2x - 3)

f(x) = 3 - 2x

When x ≥ 1.5, (2x - 3) ≥ 0.

f(x) = 2x - 3

3. Answer :

f(x) = |3 - x| + 5

Equate the expression inside the absolute value to zero and solve for x.

3 - x = 0

-x = -3

x = 3

When x ≤ 3, (3 - x) ≥ 0.

f(x) = (3 - x) + 5

f(x) = 8 - x

When x > 3, (2x - 3) < 0.

f(x) = -(3 - x) + 5

f(x) = -3 + x + 5

f(x) = x + 2

4. Answer :

f(x) = |x + 2| + |x - 3|

Equate the expression inside the absolute value to zero and solve for x.

When x < -2, (x + 2) < 0 and (x - 3) < 0.

f(x) = -(x + 2) - (x - 3)

f(x) = -x - 2 - x + 3

f(x) = 1 - 2x

When x > 3, (x + 2) > 0 and (x - 3) > 0.

f(x) = (x + 2) + (x - 3)

f(x) = x + 2 + x - 3

f(x) = 2x - 1

When -2 ≤ x ≤ 3, (x + 2) ≥ 0 and (x - 3) ≤ 0.

f(x) = (x + 2) - (x - 3)

f(x) = x + 2 - x + 3

f(x) = 5

Therefore,

5. Answer :

f(x) = |x - 5| + |x|

Equate the expressions inside the absolute value to zero and solve for x.

When x < 0, (x - 5) < 0 and x < 0.

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

f(x) = -x + 5 - x

f(x) = 5 - 2x

When x > 5, (x - 5) > 0 and x > 0.

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

f(x) = x - 5 + x

f(x) = 2x - 5

When 0 ≤ x ≤ 5, (x - 5) ≤ 0 and x ≥ 0.

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

f(x) = -x + 5 + x

f(x) = 5

Therefore,

6. Answer :

When x < 0,

When x > 0,

Therefore,

7. Answer :

Equate the expressions inside the absolute value to zero and solve for x.

x + 4 = 0

x = -4

When x < -4, (x + 4) < 0.

When x > -4, (x + 4) > 0.

Therefore,

8. Answer :

Equate the expressions inside the absolute value to zero and solve for x.

x² - 9 = 0

x² - 3² = 0

(x + 3)(x - 3) = 0

x + 3 = 0  or  x - 3 = 0

x = -3  or  x = 3

When x < -3 or x > 3, (x² - 9) > 0.

When -3 < x < 3, (x² - 9) < 0.

Therefore,

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