2
5 , which is invalid since −1
Equation 2, Case 2: 4 − | + 1| = 3 ⇒ 4 − ( + 1) = 3 ⇒ 3 − 1 = 3 ⇒
3 = 4 ⇒ =
4
3 , which is valid since ≥ −1.
Thus, the solution set is
− 2
3
4
3
.
4. | − 1| =
− 1
if ≥ 1
1 − if 1
and | − 3| =
− 3
if ≥ 3
3 − if 3
Therefore, we consider the three cases 1, 1 ≤ 3, and ≥ 3.
If 1, we must have 1 − − (3 − ) ≥ 5 ⇔ 0 ≥ 7, which is false.
If 1 ≤ 3, we must have − 1 − (3 − ) ≥ 5 ⇔ ≥
9
2 , which is false because 3.
If ≥ 3, we must have − 1 − ( − 3) ≥ 5 ⇔ 2 ≥ 5, which is false.
All three cases lead to falsehoods, so the inequality has no solution.
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CHAPTER 1 PRINCIPLESOF PROBLEMSOLVING
5. () =
2 − 4|| + 3 . If ≥ 0, then () =
2 − 4 + 3 = |( − 1)( − 3)|.
Case (i): If 0 ≤ 1, then () = 2 − 4 + 3.
Case (ii): If 1 ≤ 3,then () = −( 2 − 4 + 3) = − 2 + 4 − 3.
Case (iii): If 3, then () = 2 − 4 + 3.
This enables us to sketch the graph for ≥ 0. Then we use the fact that is an even
function to reflect this part of the graph about the axis to obtain the entire graph. Or, we
could consider also the cases −3, −3 ≤ −1, and −1 ≤ 0.
6. () =
2 − 1 −
2 − 4 .
2 − 1 =
2
− 1 if || ≥ 1
1 − 2 if || 1
and
2 − 4 =
2
− 4 if || ≥ 2
4 − 2 if || 2
So for 0 ≤ || 1,() = 1 − 2 − (4 − 2 ) = −3, for
1 ≤ || 2,() = 2 − 1 − (4 − 2 ) = 2 2 − 5, and for
|| ≥ 2,() = 2 − 1 − ( 2 − 4) = 3
7. Remember that || = if ≥ 0 and that || = − if 0. Thus,
+ || =
2
if ≥ 0
0 if 0
and + || =
2
if ≥ 0
0 if 0
We will consider the equation + || = + || in four cases.
(1) ≥ 0 ≥ 0
2 = 2
=
(2) ≥ 0, 0
2 = 0
= 0
(3) 0, ≥ 0
0 = 2
0 =
(4) 0 0
0 = 0
Case 1 gives us the line = with nonnegative and .
Case 2 gives us the portion of the axis with negative.
Case 3 gives us the portion of the axis with negative.
Case 4 gives us the entire third quadrant.
8. | − | + || − || ≤ 2 [call this inequality ()]
Case (i): ≥ ≥ 0. Then () ⇔ − + − ≤ 2 ⇔ − ≤ 1 ⇔ ≥ − 1.
Case (ii): ≥ ≥ 0. Then () ⇔ − + − ≤ 2 ⇔ 0 ≤ 2 (true).
Case (iii): ≥ 0 and ≤ 0. Then () ⇔ − + + ≤ 2 ⇔ 2 ≤ 2 ⇔ ≤ 1.
Case (iv): ≤ 0 and ≥ 0. Then () ⇔ − − − ≤ 2 ⇔ −2 ≤ 2 ⇔ ≥ −1.
Case (v): ≤ ≤ 0. Then () ⇔ − − + ≤ 2 ⇔ 0 ≤ 2 (true).
Case (vi): ≤ ≤ 0. Then () ⇔ − − + ≤ 2 ⇔ − ≤ 1 ⇔ ≤ + 1.
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CHAPTER 1 PRINCIPLESOF PROBLEMSOLVING
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Note: Instead of considering cases (iv), (v), and (vi), we could have noted that
the region is unchanged if and are replaced by − and −, so the region is
symmetric about the origin. Therefore, we need only draw cases (i), (ii), and
(iii), and rotate through 180 ◦ about the origin.
9. (a) To sketch the graph of () = max{1},