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Calculus 30 A5 – Absolute Value
A. Find, or express, the value of each of the following.
1. 25
2. 3.025
3.
52
4.
52
5. 3  2
6. 2  8
B. Rewrite each of the following absolute value expressions without absolute value signs. Begin by performing a
  , , ififxx???
sign analysis of the expression. Write your solutions in the form
??? using interval notation.

1. x  6
2. 3x  5
3. 2  10x
C. Rewrite each of the following absolute value expressions without absolute value signs. Begin by performing a
  , , ififxx???
sign analysis of the expression. Using interval notation, write your solutions in the form
??? .

1. 16  x
2
2. 18  3x  x 2
3. x 2  x  11
4. x 2  6 x  2
5.
x 5
x5
6.
x 2  25
2 x
7.
x3  27
x3
D. Use an absolute value property to explain why each of the following statements is true.
1. 5x  5 x
2.
x
x

2
2
3.
 3
4.
10
 5
2
 310
5
5. x  2  2  x
6. If x  2  4 and y  3  5 , then x  y  1  9
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E. Rewrite each of the following radicals in simplest form. Note that variables represent any real numbers.
1.
x2
2.
8b2
F. In the context of all real numbers, why is it permissible to write
G. Explain why
n
x4  x2 but incorrect to write
x2  x ?
xn  x and not x if n is an odd positive integer?
H. Rewrite each of the following expressions without absolute value signs. Begin by performing an expression
analysis. Support your solutions with interval notation.
1. x  x  6
2. x  8  x  4
3.
x4
x4
I. Solve each of the following equations.
1. 3x  7  20
2. x 2  3 x  3  1 (There are four solutions)
3. x 4  3x 2  7 x  12  6
4.
2
1

3x  4 3
J. Solve each of the following inequalities or equations. Specify your solution set.
1. 2 x  6  12 Use the shackled inequality property to write 12  2 x  6  12 . Now solve the double linear
inequality.
2. 3x  1  10 Use the here and there inequality property to write 3x  1  10 or 3x  1  10 . Solve each of these
inequalities independently and find the union (or) of the two solution sets.
3. 4  x  10
4. x  2  x
5.
x3
 0 (Note the denominator is never negative, also x  2 .)
x2
6. x  3  x  1  8
7. Find  x : x  3  1   x : x  1  3
K. Sketch the graph of each of the following absolute value functions.
1. y  3
2. y  x
3. y  2x
4. y  2 x  8
5. y  x 2  5 x  6
6. y  25  x 2
7. y  x 3
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8. y  x3  8
9. y  x3  4 x
10. y  sin x
11. y  log 2 x
L. Explain why the graphs of y  2 x and y  2 x are the same.
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Calculus 30 A5 – Absolute Value – Answer Key
A.
1. 25
2. 3.025
3.
52
4.
52
6. 8 - 2π
5. 5
B. Rewrite each of the following absolute value expressions without absolute value signs. Begin by performing a
  , , ififxx???
sign analysis of the expression. Write your solutions in the form
??? using interval notation.
1. x  6 

(xx  6),6, ififxx[6,(),6)
2. 3x  5 

3. 2  10 x 
3x  5, if x  [ 35 , )
(3x  5), if x  (, 35 )

2  10 x, if x  (, 15 ]
(2  10 x), if x  ( 15 , )
C. Rewrite each of the following absolute value expressions without absolute value signs. Begin by performing a
  , , ififxx???
sign analysis of the expression. Using interval notation, write your solutions in the form
??? .


16  x 2 , if x  [4, 4]
1. 16  x 2  
2
(16  x ), if x  (, 4) (4, )

18  3x  x 2 , if x  [3,6]
2. 18  3x  x 2  
2
(18  3x  x ), if x  (, 3) (6, )
3.
The result of attempting to
x 2  x  11  0
determine a zero value leads us
b  b 2  4ac
to a complex number, therefore
x
there is no place where the
2a
value of the expression is zero.
2
(1)  (1)  4(1)(11)
x
2(1)
Therefore the expression is
always positive or always
1  43
negative. Choose one test
2
value, 0.
Final answer is the expression is
valid for all real numbers and the
absolute value brackets are
redundant.
x2  x 11 for x 
(0)2 - (0) + 11 = 11 which is
positive, therefore all values of
the expression are positive
4. x 2  6 x  2 .
x2  6 x  2  0
x
(6)  (6) 2  4(1)(2)
2(1)
6  28 6  2 7
x

 3 7
2
2
Test values on a number line to determine
positive and negative areas
00


(3 7)
(3 7)
x2  6 x  2 
 x 2  6 x  2, if x  (, 3  7] [3  7, )

2
 ( x  6 x  2), if x  (3  7,3  7 )
Test spots
(0)2 - 6(0) + 2 = 2....positive in first interval
(2)2 - 6(2) + 2 = -4....negative in middle
interval
(0)2 - 6(0) + 2 = 2....positive in first interval
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x 5
x5
Test values on a number line to determine positive and
negative areas
5.
x 5
x  5  x  5 , if x  (, 5] (5, )

x  5  ( x  5 ), if x  (5,5)
x5

0 ( x 5)
0 ( x 5)



-5
5
0  
( x  5)
( x  5)
There is no value at 5 as it is a vertical asymptote.
x 2  25
2 x
6.
x 2  25 ( x  5)( x  5)

2 x
2 x
Test values on a number line to determine positive
and negative areas
0 ( x5)
0 ( x5)
0 (2 x)



-5
2
5
2
0 ( x 25)
(2 x )
x 2  25

2 x
 x 2  25
, if x  (, 5] (2,5]

2 x
  x 2  25 
 
 , if x  (5, 2) (5, )
  2  x 
There is no value at 2 as it is a vertical asymptote.
x3  27
x3
7.
x 3  27 ( x  3)( x 2  3 x  9)

x3
x3
Test values on a number line to determine positive and
negative areas
Note *
( x 2  3 x  9) does not factor further
0 ( x3)
0 ( x3)




-3
3
2
0 ( x 3)( x 3 x 9)
( x 3)
There is no value at -3 as it is a vertical asymptote.
 x3  27
, if x  (, 3) [3, ]
x  27  x  3

3
 x  27 
x3
 
 , if x  (3,3)

 x3 
3
D. Use an absolute value property to explain why each of the following statements is true.
1. Using |ab| = |a| |b|, then |-5x| = |-5| which leads to |x| = 5 |x|
x
x
x
a a
x

2. Using
which leads to
 , then

2
2
b b
2 2
3. Using  a   a , then  3
b
b
4. Using
a
2
 a , then
 3 which leads to  3
10
10
 5 
2
 5 which leads to
10
 5
2
 310
5
5. Using a  a then x  2  ( x  2) which leads to x  2  2  x
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6. If x  2  4 and y  3  5 , then x  2  y  3  4  5 and using the triangle inequality
x  2  y  3  x  2  y  3  9 which leads to x  y  1  9
E. Rewrite each of the following radicals in simplest form. Note that variables represent any real numbers.
1.
x2  x
2.
8b 2  2 2 b
F. In the context of all real numbers, why is it permissible to write x4  x2 but incorrect to write x2  x ?
x2 will always result in a positive value, therefore there is no need for the absolute value bars, whereas x could be
any real number including negatives, therefore | x | is necessary.
G. Explain why
n
xn  x and not x if n is an odd positive integer?
If n is odd, then there is a possibility that the value of x could be negative, for example
cube root of a negative is possible, whereas the square root of a negative is not.
3
8  3 (2)3  2 , the
H. Rewrite each of the following expressions without absolute value signs. Begin by performing an expression
analysis. Support your solutions with interval notation.
1.
Section A -x – ( - (x
x  x6
6,if x  (, 0)

– 6)) = - 6
x  2 x  6,if x  [0, 6]
  x 
 0  x 

6,if x  (6, )
Section B x – (-(x –


  ( x 6) 
 0  ( x 6) 
6)) = 2x - 6



--------------------------0----------6-----------------
A
B
C
Section C
=6
x - (x – 6)
A
2. x  8  x  4

  ( x 8)  0 
 ( x 8) 

  ( x  4) 
 0  ( x  4) 





--------------------(-4)-------------------8-----------------
A
B
C
Section A -(x –
8) – (x + 4) = -2x
+4
2 x  4,if x  (, 4)

x  12,if x  [4, 8]
2 x  4,if x  (8, )

Section B -(x –
8) + (x + 4) = 12
Section C (x –
8) + (x + 4) = 2x 4
3.
x4
x4
1,if x  (4, )
x
1,if x  (,4)
Note: 4 is not included as a vertical asymptote occurs there
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I. Solve each of the following equations.
1. 3x  7  20
2.
x 2  3x  3  1
3x – 7 = 20
3x – 7 = -20
3x = 27
x=9
3x = -13
x = -13/3
x2 + 3x – 3 = 1
x2 + 3x – 4 = 0
(x + 4)(x – 1) = 0
x = -4 x = 1
x2 + 3x – 3 = -1
x2 + 3x – 2 = 0
x
3  32  4(1)(2)
2(1)
x
3  17
2
3. x 4  3x 2  7 x  12  6 No solution as the absolute value can’t be negative.
4.
2
1

3x  4 3
2
1

3x  4 3
6  3x  4
10  3 x
10
x
3
2
1

3x  4 3
6  3 x  4
2  3 x
2
x
3
J. Solve each of the following inequalities or equations. Specify your solution set.
1. 2 x  6  12
Use the shackled inequality property to write 12  2 x  6  12 .
12  2 x  6 2 x  6  12
6  2 x
2 x  18
3  x
x9
2. 3x  1  10 Use the here and there inequality property to write 3x  1  10 or 3x  1  10 .
3 x  1  10 3 x  1  10
3x  9
3x  11
11
x3
x
3
3. 4  x  10
If x  0 , the inequality becomes 4  x  10 , which has solution x   4,10 . If x  0 , the inequality becomes
4   x  10 . Multiplying by 1 yields 4  x  10 or 10  x  4 . Thus x  10, 4 . The solution set
would be specified as  x : x   10, 4   4,10
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4. x  2  x
The absolute value expressions x  2 and x separate the number line into three regions. In the region  2,  ,
x  2  x becomes x  2  x which has no solution. In the region  0, 2 , x  2  x becomes   x  2  x
which has solution x  1 . Thus in the region  0, 2 , the solution is x   0,1 . In the region  , 0 , x  2  x
becomes   x  2   x which is true for all real numbers. Thus in the region  , 0 the solution is x   ,0 .
Uniting the two solutions we arrive at  x : x   ,1 .
5.
x3
 0 (Note the denominator is never negative, therefore x  2 .)
x2
The denominator is always positive; therefore the only concern becomes the numerator being positive, that occurs
where
x > -3, but x can’t be -2, therefore the solution is  x : x   3, 2  (2, )
6. x  3  x  1  8
There are four possible scenarios:
(x + 3) + (x + 1) = 8
(x + 3) – (x + 1) = 8
2x + 4 = 8
2=8
2x = 4
No Solution
x=2
-(x + 3) + (x + 1) = 8
-2 = 8
No Solution
-(x + 3) – (x + 1) = 8
-2x -4 = 8
-2x = 12
x = -6
7. Find  x : x  3  1   x : x  1  3
Use the shackled inequality property to write 1  x  3  1 .
1  x  3 x  3  1
2 x
x4
Use the shackled inequality property again to write 3  x  1  3 .
3  x  1 x  1  3
2  x
x4
Therefore the intersection of all four inequalities is x : x  (2,4)
K. Sketch the graph of each of the following absolute value functions.
K1
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2
3
4
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5
6
7
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8
9
10
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11
L. Explain why the graphs of y  2 x and y  2 x are the same.
y  2 x will always be above the x-axis. Whether the value of x is positive or negative the value of 2 x will ALWAYS
be positive, therefore the absolute value brackets are redundant.
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