WRITING EXPRESSIONS WITHOUT ABSOLUTE VALUE SIGNS

Using |a | = a if a ≥ 0 and –a if a < 0, write the following functions without modulus signs and hence graph each function :

Problem 1 :

y = |x – 2|

Solution :

Step 1 :

If x ≥ 2, then (x – 2) ≥ 0

If x < 2, then (x – 2) < 0, that is -(x - 2).

Step 2 :

Simplifying the second branch -(x - 2), we get

-x + 2 or 2 - x

Step 3 :

Removing the absolute sign, we get piecewise functions given below.

Problem 2 :

y = |x + 1|

Solution :

Step 1 :

If x ≥ -1, then (x + 1) ≥ 0

If x < -1, then (x + 1) < 0, that is -(x + 1).

Step 2 :

Simplifying the second branch -(x + 1), we get

-x - 1

Step 3 :

Removing the absolute sign, we get piecewise functions given below.

Problem 3 :

y = - |x|

Solution :

Step 1 :

If x ≥ 0, then y = -x

If x < 0, then y = -(-x)

Step 2 :

Simplifying the second branch -(-x), we get

y = x

Step 3 :

Removing the absolute sign, we get piecewise functions given below.

Problem 4 :

y = |x| + x

Solution :

Step 1 :

If x ≥ 0, then y = x + x

If x < 0, then y = (-x) + x

Step 2 :

Simplifying the second branch -(-x), we get

y = (-x) + x

y = 0

Step 3 :

Removing the absolute sign, we get piecewise functions given below.

Problem 5 :

y = |x|/x

Solution :

Step 1 :

If x > 0, then y = x/x

If x < 0, then y = (-x)/x

If x = 0, then y = undefined

Step 2 :

• By simplifying y = x/x, we get y = 1
• By simplifying y = -x/x, we get y = -1

Step 3 :

Removing the absolute sign, we get piecewise functions given below.

Problem 6 :

y = x - 2 |x|

Solution :

Step 1 :

If x ≥ 0, then y = x - 2x

If x < 0, then y = x - 2(-x)

Step 2 :

• By simplifying y = x - 2x, we get y = -x
• By simplifying y = x - 2(-x), we get y = x + 2x ==>y = 3x

Step 3 :

Removing the absolute sign, we get piecewise functions given below.

Problem 7 :

y = |x| + |x – 2|

Solution :

Step 1 :

By decomposing y = |x| + |x - 2| into two branches, we get

x + x – 2 and –x – (x – 2)

Step 2 :

Simplifying x + x – 2,  we get 2x – 2.

Simplifying -x -(x – 2), we get –x –x + 2, that is 2 – 2x.

Step 3 :

Problem 8 :

y = |x| - |x - 1|

Solution :

Step 1 :

By decomposing y = |x| - |x - 1| into two branches, we get

x - x – 1 and –x – (-x – 1)

Step 2 :

Then, x - x – 1, that is -1 and –x – (-x – 1) will become –x + x + 1, that is 1.

Step 3 :

Problem 9 ;

y = |x² + 1|

Solution :

Step 1 :

By decomposing |x² + 1| into two branches, we get

x² + 1 and –x² - 1

y = |x² + 1| for all x.

Problem 10 ;

y = |x² - 1|

Solution :

Step 1 :

By decomposing |x² - 1| into two branches, we get

x² - 1 and –(x² – 1)

Step 2 :

Then, -(x² – 1) will become –x² + 1, that is 1 - x².

Step 3 :

Problem 11 :

y = |x² – 2x|

Solution :

Step 1 :

By decomposing |x² – 2x | into two branches, we get

x² – 2x and –(x² – 2x)

Step 2 :

Then, -(x² – 2x) will become –x² + 2x, that is 2x - x².

Step 3 :                                                                              

Problem 12 :

y = |x² + 3x + 2|

Solution :

Step 1 :

By decomposing |x² + 3x + 2| into two branches, we get

x² + 3x + 2 and –(x² + 3x + 2)

Step 2 :

Then, -(x² + 3x + 2) that is –x² - 3x – 2.

Step 3 :