Department of Economics | Maths Revision Notes 5
1. The cubic function
General form y = ax3 + bx2 + cx + d (a, b, c, d are parameters)
1.1
Simplest case: y = x3
Fig. 7
(by setting a = 1, b = c = d = 0)
y = x3
x
-3
-2
-1
0
1
2
3
30
y
20
10
-3
-2
-1
0
1
2
y = x3
-27
-8
-1
0
1
8
27
3
x
-10
-20
-30
Key features:
(i)
(ii)
(iii)
(iv)
x
y = x3
1.2
x3 has same sign as x (that is, x3 is positive when x is positive, and negative when
x is negative).
because x3 increases/decreases very rapidly as x increases/decreases, the curve
turns sharply up/down, going off very quickly towards +∞/ -∞. (For example,
when x = +10, x3 = +1,000).
Behaviour when x is close to zero. Look at table below. Key feature is that when
x is a fraction, x3 is less than x in absolute value. So when x close to zero, y is
even closer to zero. (Note: the graph is misleading, because it appears that the
curve coincides with the axis when x close to zero. In fact, as long as x is nonzero, y is non-zero too.)
The graph of y = -x3 would be the mirror image of Fig. 7, i.e. it would slope
downward from left to right (that is, y would be positive when x was negative and
negative when x was positive).
-1
2
-1
8
-1
4
- 1
64
-1
0
1
8
- 1
0
1
512
8
512
1
4
1
64
1
2
1
8
Slightly more complex case: y = x3 + 6x2
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Key features (compared with y = x3 (above):
(i)
6x2 is positive when x is positive. So, when x positive, adding 6x2 pushes the
curve up.
(ii)
But 6x2 is also positive when x is negative. So, when x negative, adding 6x2 again
pushes the curve up. This opposes the effect of x3, which is negative when x negative and
is therefore pushing the curve down. From graph, we can see that, when x is negative but
small in absolute value, the upward pressure from 6x2 is stronger than the downward
pressure of x3, hence the curve is rising. (For example, when x = -2, x3 = -8 and 6x2 = 24,
so y = 16. See dotted lines on graph.)
Eventually, as we move leftwards and x becomes more and more negative, the downward
pressure from x3 dominates the upward pressure from 6x2, and the curve turns down. The
turning point is at x = -4. (To the left of this, the curve falls towards -∞.)
y
35
30
y = x3 + 6x2
25
20
15
Fig. 8a
10
5
0
-5
-4
-3
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-2
-1
0
1
x
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2
1.3 Adding 3x + 6
(i)
Adding the constant term, 6, just shifts the whole curve bodily upwards by 6 units.
So its y-intercept (where x = 0) is now at y = 6 instead of y = 0.
(ii)
The 3x term has the same sign as x (i.e. positive when x positive, negative when x
negative). So it pushes the curve up when x positive and down when x negative. But this
effect not very noticeable (compare Figs. 8a and 8b), because 3x does not vary with x as
much as x3 and x2 do.
Fig. 8b:
y = x3 + 6x2 + 3x + 6
y
35
30
25
20
15
10
5
-5
-4
-3
-2
-1
0
1
x
1.4 Overall conclusions.
(i) The cubic function has a characteristic
S-shape when graphed. Depending on parameter values, it may have two turning points
(as in Figs. 8a and 8b) or none (as in Fig. 7).
(ii) Again depending on parameter values, its graph may cut the x-axis once, or three
times (more on this shortly).
(iii) If the co-efficient of x3 is positive,
most parts of the graph slope are positively sloped (as in Figs. 7, 8a and 8b). If the coefficient of x3 is negative, most parts of the graph are negatively sloped.
x
-5
-4
-3
-2
-1
0
1
2
y = x3 + 6x2 + 3x + 6
16
26
24
16
8
6
16
44
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2.
Graphical solution of a cubic equation
2.1 The case with three solutions
Suppose we want to solve the cubic equation:
x3 + 6x2 + 11x + 6 = 0
3
2
We start by plotting the graph of the function: y = x + 6x + 11x + 6 (Fig. 9)
25
y
Fig. 9
20
3
2
y = x + 6x + 11x + 6
15
10
5
0
-4
-3
-2
-1
0
x
1
-5
-10
From graph, we see that curve cuts x-axis at x = -1, x = -2, and x = -3. At these
points, y = 0 (recall, y = 0 everywhere on x-axis). Therefore, x3 + 6x2 + 11x + 6 = 0 at
these points. In other words, x = -1, x = -2, and x = -3 are the solutions (roots) to the
cubic equation x3 + 6x2 + 11x + 6 = 0. (Check this by substituting each of these values
into the equation.)
It follows that the factors of x3 + 6x2 + 11x + 6 are (x + 3), (x + 2) and (x + 1). We can
check this by multiplying out:
(x + 3)(x + 2)(x + 1) = [(x + 3)(x + 2)](x + 1)
= [(x2 + 5x + 6 ](x + 1)
= x2(x + 1) + 5x(x + 1) + 6(x + 1)
= x3 + x2 + 5x2 + 5x + 6x + 6
= x3 + 6x2 + 11x + 6
All above is directly analagous to quadratic functions and quadratic equations considered
earlier, except we now have 3 roots instead of 2. By analogy with the quadratic case, you
might think a formula for solving cubic equations exists. But there is no such formula.
Instead, cubic equations can only be solved by "trial and error", or iterative, methods.
These methods involve lots of laborious calculations, nowadays done quickly by a
computer. Fortunately, we don't have to do this very often in economics.
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2.2 The case with only 1 solution
In the previous example, y = x3 + 6x2 + 11x + 6, the graph cut the x-axis 3 times (Fig. 9).
Hence the associated cubic equation had three (real) roots.
In Fig. 10 below we see the graph of y = x3 + 6x2 + 6x + 5. This cuts the x-axis only once,
at x = -5.
Therefore we can see that x = -5 is the only value of x for which y = x3 + 6x2 + 6x + 5 = 0.
So the cubic equation x3 + 6x2 + 6x + 5 = 0 has only one solution (root).1
50
Fig. 10: y = x3 + 6x2 + 6x + 5
y
40
30
20
10
0
-6
-5
-4
-3
-2
-1
1
2
-10
-20
x
-30
-40
1
This is not quite correct. It may be shown (quite easily, but we won't go into it here) that the factors
of x3 + 6x2 + 6x + 5 are (x + 5) and (x2 + x + 1). [You can check this by multiplying out.] So
x3 + 6x2 + 6x + 5 = 0 when either x + 5 = 0 or x2 + x + 1 = 0. But, x2 + x + 1 = 0 is a quadratic equation
with no real roots [check this by applying the formula]. Therefore it has two imaginary, or complex, roots
(see footnote 1 to Lecture Notes (3)). So we conclude that x3 + 6x2 + 6x + 5 = 0 has one real root (x = -5)
and two complex roots, these being the (imaginary) numbers that make x2 + x + 1 equal to zero.
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2.3 The boundary case
In the two previous cases we considered first a cubic function that cut the x-axis 3 times,
and then a cubic function that cut the x-axis only once. The boundary or “knife edge”
case which divides these two cases is where the curve just reaches the x-axis, i.e. is
tangent to it. An example is shown in Fig. 11 below, where the curve is tangent to the
axis at x = -3, and cuts the axis at x = +2.
Therefore the equation x3 + 4x2 -3x -18 = 0 has two roots, x = -3 and x = +2.
However, x3 + 4x2 -3x -18 may be factorised into (x - 2)(x + 3)(x + 3) [Check for yourself
by multiplying out]. So we say that there are three roots, x = -3 (twice!) and x = +2. This
may seem crazy, but it permits us to say that every cubic equation has three roots.
Fig. 11: y = x3 + 4x2 -3x -18
40
y
30
20
10
0
-4
-3
-2
-1
0
-10
-20
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1
2
x
3
3. The rectangular hyperbola
3.1
1
x
y=
The simplest case :
C
10
y
8
Fig 12:
y=
1
x
6
4
K
2
B
A
0
-5
-4
-3
-2
-1
0
1
2
3
4
-2
-4
-6
-8
D
-10
x
y=
-5
1
x
-4
-0.2 -0.25
-3
-2
-0.333
-1 0 1
2
-0.5 -1 ∞ 1 0.5
3
0.333
4
5
0.25 0.2
Key Features:
(i)
This function has an intriguing property. Suppose we pick any point on the curve
- say, point K, where x = ½ and therefore y = 2. Then draw a rectangle with its corner at
K and its opposite corner at the origin (shaded area). Since this rectangle has a horizontal
side of length ½ and a vertical side of length 2, its area is ½ X 2 = 1.
But, the area of this rectangle will = 1 irrespective of the position of K on the curve.
Proof: if we position K at any point where x = x0 (where of course x0 denotes any chosen
1
value of x), then the corresponding value of y will be y0 =
.
x0
Therefore the rectangle has a horizontal side of length x0 and a vertical side of length
1
1
. So its area is x0 X y0 = x0 X
= 1.
y0 =
x0
x0
Later in the course, we shall make use of this property in various economic applications.
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(ii)
The concept of a limit or limiting value.
(a) When x is positive, then as x becomes larger and larger, so y becomes smaller and
smaller. The graph gets closer and closer to the x-axis, but never reaches it (region A in
1
Fig. 12). For example, when x = 1,000,000, y =
, which is very small but
1,000,000
nevertheless positive. We can get y as close to zero as we wish, by making x sufficiently
large. But y can never actually reach zero, however large x becomes. We say that zero is
the limiting value, or limit, of , y.
We formalise this by saying:
"As x approaches infinity, so y approaches a limiting value of zero".
Lim. y
=0
x→∞
The x-axis (which is what the graph approaches as x approaches infinity) is said to be the
horizontal asymptote of the function.
And we write the above statement in mathematical notation as:
(b) In exactly the same way, we can consider what happens to the graph as x becomes
increasingly negative (region B in Fig.12). Once again we see that y approaches zero as x
approaches minus infinity. In mathematical notation we write this as:
Lim. y
=0
x → −∞
The only difference from the previous case is that y is now approaching zero from below,
i.e. y is always slightly less than zero.
(iii) We can also consider the behaviour of y as x approaches zero, while remaining
positive (region C). We see that y goes off towards +∞.
Similarly, as x approaches zero, while remaining negative (region D), we see that y goes
off towards -∞.
Thus there is no finite value of y corresponding to x = 0. For this reason, we say that this
is an example of a discontinuous function. There is a "gap" in the function at x = 0.
The y-axis (which is what the graph approaches as x approaches zero) is said to be the
vertical asymptote of the function.
In general, whenever y does something a bit funny when x approaches zero or infinity, we
call this the asymptotic behaviour of the function.
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y+a=
3.2 General case of rectangular hyperbola:
c
x+b
Here we have added 3 parameters (a, b and c) which have the effect of shifting the curve
around relative to the axes, and also changing the area of any rectangle drawn under the
curve.
Example: y + 10 = 5 (see Fig. 13). Here, a = 10, b = 2, and c = 5.
x +2
(i) From the graph, we can see that the horizontal asymptote is at y = -10. So in general
the parameter a determines the horizontal asymptote. You can check this algebraically:
in the function y + 10 = 5 , consider what happens as x → ∞. Since x is in the
x +2
5
must approach zero. But this means that y + 10 must approach zero
denominator,
x +2
too, which means that y must approach -10.
(ii) Similarly, from the graph, we can see that the vertical asymptote is at x = -2. So in
general the parameter b determines the vertical asymptote. You can check this
algebraically, by considering what must happen to y when x approaches -2.
(iii) Finally, what does the parameter c do? (In this example, c = 5). We won't prove it
here, but in fact this parameter gives the area of any rectangle (such as the shaded area in
Fig. 13), with one corner at any point K and its opposite corner at the intersection of the
asymptotes.
50
y
40
Fig. 13:
30
y + 10 =
20
10
-6
-5
-4
-3
-2
-1
0
-10
-20
-30
-40
-50
-60
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1
2
3
5
x +2
4:
Inequalities
We use inequalities frequently in economics, so it’s important to be confident about the
rules for manipulating them.
As a preliminary, we consider what it means to say that one number is greater or less than
another. We can think of numbers as being measured along a line from some arbitrary
starting point which we call zero (0); see diagram. Positive numbers are measured off to
the right, and negative numbers to the left. The line extends indefinitely far in both
directions, heading off to the left towards minus infinity (written as - ∞), and to the right
towards plus infinity (written as + ∞). Thus the number (+7), or simply 7, is measured
off as a distance of seven units to the right of zero. Similarly a negative number such as
(-3) is measured off as a distance of three units to the left of zero.
-5
-4
-3
-2
-1
0
1
2
−
3
4
5
+
Any number on the line is considered to be larger than any other number lying to the left
of it. Thus, in an obvious way, +4 is greater than +2 because +4 lies to the right of +2 on
the number line. We write this as 4 > 2, where > means "is greater than". Similarly <
means "is less than".
Slightly less obviously, -2 is greater than -4, again because -2 lies to the right of -4 on the
line. So we write -2 > -4.
Occasionally we are interested in comparing the absolute magnitude of two numbers; that
is, without paying any attention to whether they are positive or negative. We write x to
denote the absolute magnitude of some number x. Since the absolute magnitude ignores
sign, it follows that − 4 > 2, even though -4 < 2.
Notation for inequalities
x > y
means x is greater than y (this is called a “strong” inequality)
x≥y
means x is greater than, or equal to, y (a “weak” inequality)
x < y
means x is less than y
x≤y
means x is less than, or equal to, y
Double inequality: e.g. 0 < x < 1, means x is greater than zero but less than 1.
(Note 0 < x < 1 and 1 > x > 0 mean the same thing)
Rules for manipulating inequalities
With one modification, the rules for manipulating inequalities are exactly the same as the
rules for manipulating equations. That is to say, we can add or subtract anything
provided we add or subtract it to both sides of the inequality.
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Examples of adding and subtracting:
Example 1: Given 14 > 13, we can add 100 to both sides and get: 114 > 113. Or
we can subtract 5 from both sides and get: 9 > 8
Example 2: Given x > y, we can subtract y from both sides and get: x - y > 0
Similarly we can multiply or divide by anything provided we multiply or divide both
sides of the inequality by the same thing. However this is where the modification
mentioned above comes in. When we multiply or divide both sides of an inequality
by something, we have to pay attention to whether that something is positive or
negative. If negative, the direction of the inequality is reversed.
Examples of multiplying and dividing:
Example 3: given 2 > 1, we can multiply both sides by +5 and get: 10 > 5.
Example 4: given 2 > 1, we can divide both sides by +5 and get: 25 > 15
Example 5: given 2 > 1, we can multiply both sides by -5 and get: -10 < -5.
Example 6: given 2 > 1, we can divide both sides by -5 and get: − 25 < − 51
In examples 3 and 4 the direction of inequality is unchanged because we multiplied or
divided by a positive number. In examples 5 and 6 the direction of inequality is reversed
because we multiplied or divided by a negative number. That these reversals are correct
should become apparent when we think about where these numbers lie on the number
line. For example, in example 6, − 25 lies to the left of − 15 .
This means we must be very careful when multiplying or dividing an inequality. The
general approach must be that when you are multiplying or dividing an inequality by
something, and you don’t know whether that something is positive or negative, you must
open up two parallel paths of reasoning to cover the two possibilities.
Examples of inequality problems:
Example 1:
Solve the inequality -5 - 8x > 19
By "solve" is meant "find the values of x which satisfy the inequality" (i.e. the values for
which the inequality is true.)
The first step is to add 5 to both sides, giving
-8x > 24
Now we divide both sides by -8. Since this is negative, it reverses the inequality, giving
24
x <
that is, x < -3
−8
As a coarse check on this answer, we can try setting x equal to -3, then to something a bit
less then -3, then to something a bit more than -3. For example, if we set x = -3.1, then
-8x becomes -8 x -3.1 = +24.8. So the inequality becomes
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-5 + 24.8 > 19, that is
24.8 > 19 + 5 = 24
So indeed the given inequality is satisfied when x = -3.1. Since -3.1 is less than -3, this
corroborates our answer (but does not rigorously prove it correct, of course).
Example 2
for what values of x is
x2 + x > 2
First, subtract 2 from both sides, giving
x2 + x -2 > 0
The left hand side is now a quadratic expression which factorises as
(x + 2)(x - 1) > 0
We now divide both sides by (x + 2) but we need to consider two cases, according to
whether (x + 2) is positive or negative.
Case (a): Assume x + 2 > 0. This implies x > -2. Divide both sides by x + 2. After
cancelling, this leaves us with
x - 1 > 0 which is true if x > 1.
So in case (a), the inequality is satisfied (true) if x is positive and greater than +1
Case (b): Assume x + 2 < 0. This implies x < -2. Divide both sides by x + 2. This
reverses the inequality, since x + 2 is negative by assumption. After cancelling, this
leaves us with
x - 1 < 0 which is true if x < 1.
So in case (b), the inequality is satisfied (true) if x is negative and less than -2.
Collecting results, x must either be positive and greater than +1, or negative and less than
-2. In symbols, x > 1 or x < -2. (See next page for alternative graphical solution).
Alternative method. given (x + 2)(x - 1) > 0 from above, we see that the left hand
side is positive only if (x+2) and (x - 1) have the same sign (that is, either both positive or
both negative). After some thought we see that they are both positive if x > +1 and both
negative if x < -2. This is a quicker way of answering the question, but it's less reliable
because it relies more on intuition.
----------------------------------------------------------------
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Example 3 (Difficult, so skip it if you wish)
1 1
>
?
a b
To answer this fully we we have to consider all four possibile combinations of signs for a
and b: (i) a > 0, b > 0 : (ii) a < 0, b < 0 : (iii) a > 0, b < 0 : (iv) a < 0, b > 0
For what values of a and b is it true that
Case (i) a > 0, b > 0
Step 1. Multiply both sides by a. Since a > 0, this leaves the direction of inequality
a
unchanged, and gives 1 >
b
Step 2. Multiply both sides by b. Since b > 0, this leaves the direction of inequality
unchanged, and gives gives b > a
1 1
So from case (i) we conclude that > is true when a > 0 and b > 0 provided b > a.
a b
Case (ii) a < 0, b < 0
Step 1. Multiply both sides by a. Since a < 0, this reverses the direction of inequality,
a
and gives 1 < . Step 2. Multiply both sides by b. Since b < 0, this reverses the
b
direction of inequality, and gives b > a
1 1
So from case (ii) we conclude that > is true when a < 0 and b < 0 provided b > a.
a b
Case (iii) a > 0, b < 0
Step 1: Multiply both sides by a. Since a > 0, this leaves the direction of inequality
a
unchanged, and gives 1 > . Step 2: Multiply both sides by b. Since b < 0, this reverses
b
1 1
the direction of inequality, and gives b < a. So from case (iii) we conclude that > is
a b
true when a > 0 and b < 0 provided b < a. But the restriction b < a is redundant , since if
a is positive and b negative (as assumed in this case), it follows automatically that b is
less than a (since any negative number is smaller than any positive number)
Case (iv) a < 0, b > 0
Step 1: Multiply both sides by a. Since a < 0, this reverses the direction of inequality
a
and gives 1 < . Step 2: Multiply both sides by b. Since b > 0, this leaves the direction
b
of inequality unchanged, and gives b < a. But this is a contradiction. For if a is negative
and b positive (as assumed in this case), b < a is impossible since any positive number is
1 1
larger than any negative number. So we conclude that > is never true when a < 0
a b
and b > 0.
1 1
Collecting results we can say that > is true when a and b are both positive or both
a b
negative, provided b > a (cases (i) and (ii)). It is also true when a is positive and b
negative, irrespective of the relative magnitudes of a and b (case (iii)). It is never true
when a < 0 and b > 0 (case (iv)).
-----------------------------------------------------------
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Graph for Inequalities Example 2
4
y
3
2
1
0
-3
-2
-1
1
x
2
-1
-2
-3
Graph is of the function y = x2 + x - 2. We can see from the graph that y > 0 when x < -2
or x > 1; that is, to the left of the left-hand dotted line, or to the right of the right hand
dotted line.
Since y = x2 + x - 2, it follows that when y > 0, x2 + x - 2 is also > 0.
So we conclude that x2 + x - 2 > 0 when x < -2 and x > 1.
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5:
The circle and the ellipse (Optional topic, not included in revision programme
and will not be tested in exams)
5.1
The function: x2 + y2 = c2
(where c is a parameter) is the equation of a circle
with its centre at the origin and radius c.
Proof: Draw a circle centred at the origin with radius c. Let A be any point on the circle
(see Fig. 14). Then AB = y0 = the y co-ordinate of A. Similarly 0B = x0 = the x-coordinate of A.
Since 0AB is a right-angled triangle, then by Pythagoras's Theorem 0A2 = 0B2 + AB2.
But 0B2 = x02 and AB2 = y02. Also 0A2 = c2 by construction.
Therefore at point A (with coordinates x0 and y0), x02 + y02 = c2.
But this is true of every point on the circle (since A was chosen arbitrarily). So every
point on the circle satisfies the condition x2 + y2 = c2.
y
Fig. 14:
Graph of x2 + y2 = 100
A
5
.5
B
-15
-10
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-5
0
5
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10
15
x
5.2
Similarly we can show that (x + a)2 + (y + b)2 = c
is the equation of a circle of radius c, with its centre at the point x = -a, y = -b.
Fig. 15: graph of (x + a)2 + (y + b)2 = c2
y
0
x
(-a, -b)
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5.3 The ellipse.
We can turn the circle x2 + y2 = c2 into an ellipse by adding the term xy. So the equation
of an ellipse is x2 + xy + y2 = c2.
When x and y are both positive or both negative, their product xy is positive, so the values
of x and y which satisfy x2 + xy + y2 = c2 are smaller than those which satisfy x2 + y2 = c2.
So the ellipse (dotted line) lies inside the circle. (Quadrants A and C)
When x and y have opposite signs, their product xy is negative, so the values of x and y
which satisfy x2 + xy + y2 = c2 are larger than those which satisfy x2 + y2 = c2. So the
ellipse (dotted line) lies outside the circle. (Quadrants B and D)
On the axes, either x or y is zero, so the product xy is zero too. The circle and ellipse
therefore coincide.
You will encounter the ellipse in Econometrics in your 2nd year.
Fig. 16
B
A
D
C
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