Further Concepts for Advanced Mathematics - FP1
Unit 3 Graphs & Inequalities – Section3c Inequalities
Types of Inequality
An inequality is a mathematical statement containing one of the symbols < , > , ≤ or ≥ .
There are two types of inequality:
Inequalities whose truth depends on the value of the variable concerned e.g. x + 3 > 5 is
only true if x is greater than 2.
Inequalities that are always true e.g. x 2 ≥ 0 is always true for real values of x .
The work in this section covers the first type of inequality.
Rules for manipulating inequalities
1. You may add or subtract the same value from both sides of the inequality
e.g. x > 8 ⇔ x + 3 > 8 + 3 The symbol ⇔ means “implies that”
2. You may multiply or divide both sides of an inequality by the same positive number
e.g. x > 8 ⇔ 3 x > 3 × 8
3. If both sides of an inequality are multiplied or divided by a negative number, the
inequality is reversed
e.g. x > 8 ⇔ −3 x < −3 × 8
4. You may add (but not subtract) corresponding sides of inequalities of the same type
e.g. if x > 8 and y > 5 then x + y > 8 + 5
5. Inequalities of the same type are transitive.
e.g. if x > y and y > z then x > z
You cannot subtract corresponding sides of inequalities of the same type. This example
shows why it should be avoided.
If x > 8 and y > 5 then x − y > 8 − 5 giving x − y > 3
But what if x = 10 and y = 12 ? These values are true for the initial inequalities but
clearly show that the subtraction does not work.
1
Using sketch graphs to solve inequalities
Examples
Solve the inequality
5x − 2
≥0
( x + 1)( x − 2)
First sketch the graph of y =
When x = 0 , y =
−2
=1
1 × −2
When y = 0 , x =
2
5
5x − 2
( x + 1)( x − 2)
Vertical asymptotes are x = −1 and x = 2
As x → −1 from the left, y → −∞
As x → −1 from the right, y → +∞
As x → 2 from the left, y → −∞
As x → 2 from the right, y → +∞
As x → −∞ , y →
5x 5
5
= =
= 0 from the negative side
2
x −∞
x
As x → +∞ , y →
5x 5 5
= = = 0 from the positive side
x2 x ∞
The sketch should look like this:
y
We want the values of x for which the
y values are ≥ 0
1
-1
2
5
2
x
2
These are shown on the sketch on the
next page.
The values for which y ≥ 0 are marked
by the horizontal lines in the graph.
y
1
-1
2
5
2
x
The solution to the inequality is therefore − 1 < x ≤
The filled circle denotes where a
value exists and may be used in the
inequality. The empty circles denote
values that are either undefined (as
both are in this case) or are not true
in the inequality.
2
5
or x > 2
An algebraic method
5x − 2
≥ 0 all we need to do is think in terms of positive and
( x + 1)( x − 2)
negative numbers since ≥ 0 means either positive or 0.
To solve the inequality
Think about the critical points for each part of the function
2
2
since values of x lower than
will give a negative
5
5
result and higher values will give a positive result.
For 5 x − 2 the critical point is
For x + 1 the critical point is -1
For x − 2 the critical point is 2
We can put these on a number line and then consider where
negative.
3
5x − 2
is positive or
( x + 1)( x − 2)
2
5
-1
2
5x − 2
-ve
-ve
+ve
+ve
x +1
-ve
+ve
+ve
+ve
x−2
-ve
-ve
-ve
+ve
-ve
+ve
-ve
+ve
5x − 2
( x + 1)( x − 2)
The required regions are clear from the table. The main thing to get right here is that
the function is undefined for the x values -1 and 2.
The table gives the same solution − 1 < x ≤
2
5
or x > 2
Inequalities of the form f ( x) ≤ g ( x)
There are two ways to solve these:
1. By drawing the graphs of both y = f ( x) and y = g ( x) and looking to see where one is
lower than the other.
2. By rearranging the inequality to f ( x) − g ( x) ≤ 0 and then drawing the graph of
y = f ( x) − g ( x) .
Example
Solve the inequality 8 − x ≥
x+2
x−2
Method 1
The sketch graph of y = 8 − x is simple to draw.
For y =
x+2
a little more effort is needed.
x−2
2
= −1
−2
Vertical asymptote: x = 2
Intercepts: x = 0 , y =
y = 0 , x = −2
As x → 2 from the left, y → −∞
as x → 2 from the right, y → +∞
4
Behaviour as x → ±∞ :
As x → −∞ , y → 1 from below
The two sketch graphs look like this:
as x → +∞ , y → 1 from above
y
y=
x+2
x−2
x
y = 8− x
The inequality 8 − x ≥
y=
x+2
.
x−2
x+2
is satisfied where the graph of y = 8 − x is above the graph of
x−2
The set of values for x that satisfy the inequality are marked on the diagram.
One inequality includes both points of intersection so there are filled circles at each end
of the line. The other does not include x = 2 as the function is undefined for that value
and so an empty circle has been used.
We have to find the points of intersection to solve the inequality fully:
x+2
x−2
= x+2
= x+2
= x 2 − 9 x + 18
=0
8− x =
(8 − x)( x − 2)
− x 2 + 10 x − 16
0
( x − 3)( x − 6)
5
So x = 3 or x = 6
The solution to the inequality 8 − x ≥
x+2
is 3 ≤ x ≤ 6 or x < 2
x−2
Method 2
x+2
≥0
x−2
Rearrange the inequality to
8− x−
Using a common denominator
(8 − x)( x − 2) − ( x + 2)
≥0
x−2
multiplying out
− x 2 + 10 x − 16 − x − 2
≥0
x−2
and simplifying gives
− x 2 + 9 x − 18
≥0
x−2
This looks a little unpleasant so to make the x 2 term positive, we multiply both sides by
-1 which reverses the inequality
x 2 − 9 x + 18
≤0
x−2
( x − 3)( x − 6)
≤0
x−2
Factorising the top gives
To solve the inequality (and hence our original one), we need to draw the graph of
( x − 3)( x − 6)
y=
and find where the y values are negative (i.e. <0)
x−2
The sketch graph of y =
Intercepts: x = 0 , y =
( x − 3)( x − 6)
x−2
18
= −9
−2
y = 0 , x = 3 or x = 6
Vertical asymptote x = 2
As x → 2 from the left, y → −∞
as x → 2 from the right, y → +∞
Behaviour as x → ±∞ :
x2
x2
As x → −∞ , y →
= x = −∞
as x → +∞ , y →
= x = +∞
x
x
There are no horizontal asymptotes (there is an oblique asymptote – can you work out
what is is?).
6
The sketch graph looks like this:
y
2
3
6
x
From the graph you can see that the solution the the inequality
3 ≤ x ≤ 6 or x < 2
Hence the solution to the inequality 8 − x ≥
( x − 3)( x − 6)
≤ 0 is
x−2
x+2
is 3 ≤ x ≤ 6 or x < 2
x−2
Solving the inequality algebraically
Rearrange the inequality 8 − x ≥
method.
x+2
( x − 3)( x − 6)
to
≤ 0 as we did for the second
x−2
x−2
The critical points are x = 2 for x − 2 , x = 3 for x − 3 and x = 6 for x − 6 .
7
Putting these on a number line gives:
3
2
6
x−2
− ve
+ ve
+ ve
+ ve
x−3
x−6
− ve
− ve
− ve
− ve
+ ve
− ve
+ ve
+ ve
− ve
+ ve
− ve
+ ve
( x − 3)( x − 6)
x−2
So the inequality is true for values of x that are less than 2 and values of x between 3
and 6.
We need to check if, for any of these values, the function is undefined.
For x = 2 , the function
( x − 3)( x − 6)
is undefined as the denominator would be 0.
x−2
This means that we don’t include x = 2 in our final solution.
The table shows that the solution to the inequality 8 − x ≥
(yet again!)
8
x+2
is 3 ≤ x ≤ 6 or x < 2
x−2