Download Chapter 2 Inequalities

Chapter 2
Inequalities
After the completion of the lectures and tutorials associated with this chapter you should
be able to:
• Understand the basic rules of order in the real numbers.
• Use these rules to solve simple inequalities including quadratics and those
involving the modulus function.
TEXTBOOK – Adams P1 - P3 are relevant here.
2.1
Basic Inequalities
Real numbers have a property of order or size. When the real number a is greater than
or equal to the real number b, we write a ≥ b. When x is smaller than y, we write x < y
etc.
Important properties are:
• a > b ⇐⇒ a + c > b + c
∀a, b, c ∈ R
• a > b and c > 0 ⇐⇒ ac > bc
• a > b and c < 0 ⇐⇒ ac < bc
NOTE There are equivalent versions with ‘>’ replaced by ‘≥’ etc.
A consequence of these properties is that for example:
ab > 0 ⇐⇒ [(a > 0) and (b > 0)] or [(a < 0) and (b < 0)]
1
2
CHAPTER 2. INEQUALITIES
2.2
Solving Inequalities
These rules allow us to solve inequalities.
Inequalities generally take the form:
p(x) > 0
where p(x) is some expression in an unknown quantity x. The ‘<’ sign could be replaced
by ‘≥’, ‘<’ or ‘≤’.
The object is usually to determine the solution set
S = {x : p(x) > 0 is true}
This involves showing two things:
• If p(x) > 0 then x ∈ S
• If x ∈ S then p(x) > 0.
Determining S means expressing it as the union of intervals etc.
I Example 15 J
Solve x2 > 9
Claim solution set is
{x ∈ R : (x < −3) or (x > 3)}
Proof Suppose x2 > 9
∴
∴
∴
∴
x2 − 9 > 0
(x − 3)(x + 3) > 0
[(x − 3) > 0 and (x + 3) > 0] or [(x − 3) < 0 and (x + 3) < 0]
x ∈ (−∞, −3) ∪ (3, ∞)
Conversely, if x ∈ (−∞, −3) ∪ (3, ∞) then x2 > 9 (check this).
So this is the solution set.
NOTE It is assumed that all readers can solve such quadratic inequalities. However,
more subtle examples require real care.
2.2. SOLVING INEQUALITIES
3
√
I Example 16 J Solve 2x + 3 > x.
√
√
NOTE · ALWAYS means the positive square root, and for x to
exist we must have x ≥ 0.
“Solution”
√
2x + 3 > x
2x + 3 > x2
?? DANGER ??
∴ 0 > x2 − 2x − 3
∴ 0 > (x + 1)(x − 3)
∴ x ∈ (−1, 3)
NOTE Squaring on both sides requires real care. Is is only valid if both
sides are known to be ≥ 0.
This “answer” above is not quite correct. The way forward is to use two
cases:
(a) x ≥ 0,
(b) x < 0.
Part (a) Since both sides are ≥ 0 then argue as above to obtain (−1, 3)∩
[0, ∞) = [0, 3).
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Part (b) When x < 0 then we only have x ≥ − available. Here we
2
√
3
have only − ≤ x < 0, and for these x, in 2x + 3 > x, LHS ≥ 0 and
2 £
¢
RHS < 0, so ∴ − 32 , 0 is in the solution set.
So the solution set is
·
¶
·
¶
3
3
− , 0 ∪ [0, 3) = − , 3 .
2
2
We must now check that every member of this set satisfies the
inequality (exercise).
NOTES
(i) Take care when squaring.
(ii) Take care when “cross-multiplying”, in fact avoid using cross multiplying if possible.
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CHAPTER 2. INEQUALITIES
I Example 17 J
Solve
2
x+3
>
x+3
2
?? CANNOT multiply both sides by (x + 3) and 2 to give 4 > (x + 3)2 ??
Why?
Solution
Firstly, note that x = −3 is excluded since the LHS is not defined. We
have
2
x+3
>
x+3
2
∴
∴
∴
∴
x+3
2
−
x+3
2
4 − (x + 3)2
2(x + 3)
(2 − (x + 3))(2 + (x + 3))
2(x + 3)
(−x − 1)(x + 5)
2(x + 3)
> 0
> 0
> 0
> 0
Which gives (check this):
solution set = (−∞, −5) ∪ (−3, −1)
2.3. INEQUALITIES INVOLVING THE MODULUS FUNCTION
2.3
Inequalities Involving the Modulus Function
Definition 2.1 the modulus function, denoted by |x| is defined by:.
(
x
: when x ≥ 0
|x| =
−x : when x < 0
2.3.1
Basic Properties of the Modulus Function
(i) |x| = | − x| for all x ∈ R
(ii) |a − b| = |b − a| for all a, b ∈ R
(iii) |x|2 = x2 for all x ∈ R
(iv) |x| < b ⇐⇒ −b < x < b where b > 0 (there is a similar version for ≥)
I Example 18 J
Solve
|x + 2| > |x − 1|
∴
∴
∴
∴
|x + 2|2
(x + 2)2
x2 + 4x + 4
6x
∴
x
|x − 1|2
(x − 1)2
x2 − 2x + 1
−3
1
> −
2
>
>
>
>
1
Now check that every x ∈ (− , ∞) is a solution.
2
NOTE In the first stage of the solution it is okay to square both sides
of the equation as each side is ≥ 0.
If in doubt you can go back to the definition and split the problem into cases:
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CHAPTER 2. INEQUALITIES
I Example 19 J
Solve
|x2 − 40| < 24
Using property (iv) of the modulus function, we have:
|x2 − 40| < 24
⇐⇒
Suppose
∴
∴
−24 < x2 − 40 < 24
−24 < x2 − 40 < 24
16 < x2 < 64
x ∈ (−8, −4) ∪ (4, 8)
Now check that every element of this set is a solution.
2.4
Proving inequalities
Sometimes it is necessary to prove that an inequality holds for a range of values of its
variable(s).
I Example 20 J
Prove that (n + 1)3 < 2n3 for all n ≥ 4.
Proof:
(n + 1)3
now 3n + 1
∴ (n + 1)3
also 4n
∴ (n + 1)3
finally 4n2
∴ (n + 1)3
=
<
<
≤
<
≤
<
n3 + 3n2 + 3n + 1
4n (since n > 1)
n3 + 3n2 + 4n
n2 (since n ≥ 4)
n3 + 4n2
n3
n3 + n3