linear relationships nonlinear relationships linear · pdf fileec1008 handout 1: functions and...
TRANSCRIPT
EC1008 Handout 1: FUNCTIONS AND GRAPHS
• Linear Relationships • Nonlinear Relationships
Linear Equations
Y = a + bX a, b positive parameters
Key feature: slope is not a function of position on the line (i.e., not a function of the value of X).
Y
X
a b
Nonlinear Relationships
Key difference between linear and nonlinear relationships: slope becomes a function of X. Four important nonlinear relationships: • Quadratic functions • Power functions • Logarithmic functions • Exponential functions
Quadratic Function
2cXbXaY +++= a, b, c positive parameters
Y
Power Functions
Y = a + bXc, a, b, c parameters. Suppose a, b, c positive, c > 1.
Exponential Function
Y = exp(X) = eX, e ≈ 2.7183
Y
X
a
X
eX
X
Logarithmic Function
Y = ln(X)
Summary
Generic representation of a function: Y = f(X) Five important examples: linear functions: bXaY += quadratic functions 2cXbXaY +++=
power functions: CbXaY += exponential function: ))(exp( XfY = logarithmic function: ))(ln( XfY = Logarithmic transformations: ,aXYGiven b= ).Xln(b)aln()Yln( •+= ),Xexp(aYGiven •= .X)aln()Yln( +=
ln(X)
X 1
A horizontal line has zero slope
0
10
20
30
0 5 10x
yy = 18
slope = 0
as x increases, y does not change
Positive slope, zero intercept
0
250
500
0 25 50x
y
y = 9xas x increases,
y increases
slope = 9
line passes through the origin
Negative slope, positive intercept
0
10
20
30
40
50
60
0 5 10 15x
y
y = 50 - 4x
larger x values go with smaller y values
slope = - 4
Positive slope, negative intercept
-30-20-10
010203040
x
y
10 20
y = -25 + 3x
line cuts y axis below the origin
slope = 3
as x increases, y increases
A vertical line has infinite slope
0
10
20
30
40
0 5 10 15 20x
y x = 15y increases but x does not change
slope = ∞
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Table 2.3 Demand scheduleQuantity (x) Price (y)0 10040 8080 60120 40160 20200 0Intercept (vertical) = 100Slope = - 0.5: Horizontal intercept = 200
Demand Function P =100 - 0.5Q :
Join the points and label the graph
P =100 - 0.5Q
Quantity Q
PriceP
: Demand
P = 100 - 0.5Q
D
0
20
40
60
80
100
120
0 40 80 120
160
200
240
Figure 2.17
2Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
The General Linear Demand function such as
0
20
40
60
80
100
120
0
40 80 120
160
200
240
Figure 2.19 Demand function, P = 100 - 0.5Q
a = 100
P = 100 - 0.5Q
D
Q
P
Slope = - b = - 0.5
ab= 200
P = a - bQ P =100 - 0.5Q
3Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Supply Function P = 10 + 0.5QCalculate and plot the supply schedule P = 10 + 0.5Q
0
10
20
30
40
50
60
70
-20 0 20 40 60 80 100
Table 2.4 Supply scheduleQ: Quantity P: P = 10 +0.5Q0 10: P =10 +0.5(0)20 20: P = 10+0.5(20)40 3060 4080 50100 60Intercept (vertical) = 10Slope = 0.5Horizontal intercept = - 20
P = 10 + 0.5Q
NOTE: the supply function may be plotted by simply joining the intercepts
Figure 2.22
c = 10
P = 10 + 0.5Q
Slope = 0.5
Q
P
S
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
The equation of a line
Putting it another way:the equation of a line may be described as the formula that allows you to calculate the y co-ordinate for any point on the line, when given the value of the x co-ordinate.
Example: y = x is a line which has a slope = 1, intercept = 0
Example:y = x + 2 is the line which has a slope = 1 , intercept = 2
The equation of a line may be written in terms of the two characteristics, m (slope) and c (intercept) . y = mx + c
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
How to plot any Linear Budget Constraint
Rearrange the equation in the form y = mx + c (see above)Plot y on the vertical axis, against x on the horizontal axisCalculate and plot the vertical and horizontal interceptsJoin the points and label the graph
xP yP M y MP
PP
xX YY
X
Y+ = → = −
⎛
⎝⎜
⎞
⎠⎟
MP X
MPYQuantity of good Y , y
Quantity of good X, x0
10
20
30
40
0 30 60 90
2Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Plot the Budget Constraint: 2x + 6y = 180where PX =£2: PY = £6: M = 180: x(2) + y(6) = 180
Plot the horizontal intercept: x = 90Plot the vertical intercept: y = 30Join these points
0
10
20
30
40
0 30 60 90
Quantity of good Y , y
Quantity of good X, x
y = 30 - 0.33 x
MP X
MPY
Slope = −PP
X
Y
Figure 2.39
3Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
1) If Px = 3, PxX+PyY=M: 3x+ 6y=180 : y=30-0.5x2) Now adjust the graph of the Budget Constraint(y=30 - 0.5x) when the price of good X decreases
Intercept is the same: Slope has changed from -0.5 to -0.25
Figure 2.40 ∆PX and its effect on the Budget constraint
0
10
20
30
0 20 40 60 80 100 120
y = 30 - 0.5x Original Constraint
x
y
y = 30 - 0.25x Constraint after P changed
4Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Adjust the graph of the Budget Constraint: y = 30 - 0.5xwhen the per unit price of good Y decreases from 6 to 3
The adjusted constraint pivots upwards from the unchanged horizontal intercept (see Figure 2.41)Comment: When Y decreases in price, more units of Y are affordable
Figure 2.41 ∆Py and its effect on the budget constraint
0
10
20
30
40
50
60
70
0 20 40 60
y = 30 - 0.5 x
y = 60 - x
x
y
5Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Summary: Change in the graph of the Budget Constraint:y = 30 - 0.5x when the budget limit increases
Slope is the same: intercept has changed from 30 to 40When the budget limit increases, the constraint moves upwards, parallel to the original constraintComment: When the budget limit increases, more units of both X and Y are affordable
Figure 2.42 ∆Y and its effect on the Budget constraint
0
10
20
30
40
50
0 20 40 60 80x
y
Budget = 240
Budget = 180
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Quadratic Function
The basic quadratic is y = x2
The quadratic, y = ax2 : (a > 0) is wider than y = x2 when a < 1:is narrower than y = x2 when a > 1:
A quadratic equation takes the form ax2 + bx + c = 0
You can solve it graphically or by using the formula
aacbb
242 −±−
=x
2Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Worked Example 4.4(a)y = 0.6 x2 is wider than y = x2Worked Example 4.4(a)y = 0.6 x2 is wider than y = x2
5
15
25
35
-4 -3 -2 -1 0 1 2 3 4 x
y
Figure 4.5
y = x2
y = 2x2
y = 0.6x2
x -3 -2 -1 0 1 2 3
y =0.6x 2 5.4 2.4 0.6 0 0.6 2.4 5.4
3Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Worked Example 4.3 The graph of y = - x2 is the reflection of the graph of y = x2 through the x-axis
x -4 -3 -2 -1 0 1 2 3 4y = x2 16 9 4 1 0 1 4 9 16y = -x2 -16 -9 -4 -1 0 -1 -4 -9 -16
y = - x 2-20-15-10-5
5101520
-4 -3 -2 1 2 3 4x
y y = x2
Figure 4.4
4Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Properties of quadratic functions, illustrated graphically
If c<0 intercept is below x-axis, if c>0 aboveThe quadratic is a minimum type if a > 0, a maximum type if a < 0The graph is symmetrical about the vertical line drawn through the maximum or minimum pointThe roots are at the points of intersection with the x-axisThe roots are equidistant, (one greater, one smaller), from the x-coordinate of the maximum or minimum point
These point are illustrated in Worked Example 4.5, for the quadratic: y = 2x2 - 7x -9
5Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Measure the distance (along the horizontal axis) between the roots and the vertical line thro’ the minimum point: y = 2x2 - 7x + 9
2.752.75root = 4.5root = -1
(1.75, - 17)Minimum point
Figure 4.6 Graph for Worked Example 4.5
x
-17
-9
9
18
-4 -3 -2 -1 0 1 3 4 5 6
y = 2 x 2 - 7 x - 9
y
6Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Find the roots of 2x2 - 7x - 9 = 0using the ‘-b’ formula
2x2 - 7x - 9 = 0
Thus, confirming the answer obtained graphically
x = − − ± − − −( ) ( ) ( )( )( )
7 7 4 2 92 2
2
=± +7 49 72
4
4=
±=
±7 121 7 114
4 4x = +
= =7 11 18 4.5 x =
−=−
= −7 11
44
41or
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Total Revenue TR = 50Q - 2Q2:: Summary
From the graph, the maximum TR = 312.5, when Q = 12.5From the graph, TR = 0 at Q = 0 and Q = 25 (roots)
Roots
Figure 4.8 Total revenue function TR = 50Q - 2Q2
0
120
240
360
0 5 10 15 20 25
TR= 312.5
Q =12.5
30
Maximum
Q
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Rule 1
To multiply numbers with
the same base, add the
indices.
a m × a n = a m + n 5 2 × 5 3 = 5 2 + 3 = 5 5 ...by rule
(5 × 5) × (5 × 5 × 5) = 5 5 ...directly
Rule 2
To divide numbers with the
same base, subtract the
indices.
aa a
m
nm n= − 3
33 3
3 3 3 33 3
3 31
3
4
24 2 2
2
= =
× × ××
= × =
− . . .
. . .
by rule
directly
Rule 3
To raise an exponential to
a power, multiply the
indices.
( ) ka am m k= × ( ) by rule
(2 2 2)(2 2 2) = 2 directly
2
6
2 2 23 3 2 6= =
× × × ×
× . . .
. . .
Rules for indices
2Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Converting from Index to Log
for example:
(b) 8 23= , then
log 2 (8) = 3
(c)
0 125 8 1. = = −− then log (0.125) 18
Number basepower=
Number basepower= the base of the index becomesthe base of the log, the power
then log (Number) powerbase =drops down.
So, logs are powers!
3Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Rules for LogsOperation Notation Example
Rule 1Add
log b M + log b N ⇔ log b MN ln(4) + ln(29) = ln(4 × 29)
1.3863 + 3.3673 = ln(116)
4.7536 = 4.7536
Rule 2Subtract
log b M - log b N ⇔ log bMN
⎛⎝⎜
⎞⎠⎟ log(90) - log(26) = log 90
26⎛⎝⎜
⎞⎠⎟
1.9542 - 1.4150 = log(3.4615)
0.5392 = 0.53926
Rule 3Log of an
exponentiallog b(M z) ⇔ z log b (M )
log(5 3) = 3 log(5)
log(125) = 3(0.69897)
2.09691 = 2.09691
Rule 4Change of base log b
x
x
NNb
( )log ( )log ( )
⇔
the new base, x , is usually
10 or e since both are readily
available on the calculator.
log ( ) log( )log( )
.
..2 16 16
21 20410 3010
4 0= = =
or change to base e
log ( ) ln( )ln( )
.
..2 16 16
22 77260 6931
4 0= = =
4Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Antilogs: Going from log form to index form
Number basepower=
Number basepower= the base of the power becomes
then log (Number) powerbase = the base of the log, the power
drops downthen, taking antilogs is described as reversing the the process of taking logs,
logbase (Number) power= ...the base of the log becomes
(Number) basepower= the base of the power, the power goes up.
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Figure 4.15: As the size of the base increases, the graph becomes steeper For example, base a increases from a = 2 to a = e (2.718..) to a = 3.5,
0
1
2
3
4
5
6
-3 -2 -1 0 1 2 3
y = ( 2 ) x
y = ( 3 . 5 ) x y = e x
x
y
Figure 4.15 Graphs for Tables 4.10 and 4.11
2Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
The graphs of ex and e-x: summary
From the Table: when x is replaced by -x, points for e-x are reflected to ex and vice-versa
y = ex
The sketch below confirms that the graphs are mirror images through the y-axis
x -2.00 -1.50 -1.00 -0.50 0.00 0.50 1.00 1.50 2 2.5
0.14 0.22 0.37 0.61 1.00 1.65 2.72 4.48 7.39 12.18
7.39 4.48 2.72 1.65 1.00 0.61 0.37 0.22 0.14 0.08y = e-x
-2.0
0
-1.5
0
-1.0
0
-0.5
0
0.00
0.50
1.00
1.50
y = ex
y = e-x
2.00
4.00
6.00
8.00
10.00
12.00
14.00
2
2.5x
(growth curve)(decay curve)
1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Worked Example 4.15(b) Algebraic solution
The population in any year is calculated from the equation
where t = 0 at the beginning of 1980
To calculate the population in 1990, (a) determine the value of t: t = 1990 - 1980 = 10(b) substitute t = 10 into the equation for population
= 1016.4 (million)Similarly, in 2000 (a) t = 20 (b) Population = 1372 (million)
P e t= 753 0 03.
P e= 753 0 03 10. ( )
2Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd
Worked Example 4.15(c)
Start with the population of 1750, on the vertical axisDraw a horizontal line across to the graphFrom this point on the graph, drop a vertical line down to the horizontal axisRead off the year. The population is 1750 in the year 2008 approximately
250
2250
1980 1985 1990 1995 2000 2005 2010
1750
753
population =
Year
FUNCTIONS AND GRAPHS: SUMMARY
Let pxy =
If 1−=p 1−= xy exponential function
If 0=p 1=y linear function (b=0)
If 1=p xy = linear function (a=0)
If 2=p 2xy = quadratic function
• Functions of One Variable bxay +=
2cxbxay ++= xey =
• Functions of Two (or more) Variables
czbxay ++=
Functions of More Than 1 Variable
• Multivariate function: the dependent variable, y, is a function of more than one independent variable
• If y = f(x,z) y is a function of the two variables x and z
• We substitute values for x and z to find the value of the function
• If we hold one variable constant and investigate the effect on y of changing the other, this is a form of comparative staticsanalysis
Total and Average Revenue
• TR = P.Q• AR = TR/Q = P
• A downward sloping linear demand curve implies a total revenue curve which has an inverted U shape
• Symmetric: the shape of one half of the curve is the mirror image of the other half
Total and Average Cost
• Total Cost is denoted TC • Fixed Cost: FC is the constant term in TC• Variable Cost: VC = TC – FC• Average Cost per unit output:
AC = TC/Q• Average Variable Cost:
AVC = VC/Q• Average Fixed Cost:
AFC = FC/Q
Profit
• Profit = π = TR – TC• If TC = 120 + 45Q – Q2 + 0.4Q3
• and TR = 240Q – 20Q2
• π = TR – TC substitute using brackets • π = 240Q – 20Q2 – (120 + 45Q – Q2 + 0.4Q3)
Taking the minus sign through the brackets and applying it to each term in turn gives
• π = 240Q – 20Q2 – 120 – 45Q + Q2 – 0.4Q3
and collecting like terms we find
• π = – 120 + 195Q – 19Q2 – 0.4Q3