MUSTAPHA MUKTAR, PhD
Department of Economics,
Bayero University, Kano
Introduction
A function is described as a sign of expressing relationship between a dependent variable and one or more independent variables. A function merely represents a rule which specifies a particular type of relation between two or more variables. If a relation between two real variables x and y is such that when x is given y is determined, then y is said to be a function of x, this is usually expressed as;
y = ¦ (x)
and read “y equals to a function of x”, where ¦ is some kind of rule governing the qualitative (quantitative) inter-dependence of the variables y and x. As a functional notation to express the idea of one variable’s dependence on another, ¦ is read “is a function of “and means “depends upon”. The expression given above therefore means that y depends upon or varies with x, whatever y and x may be. In effect, the two variables x and y do not change independent of each other, but rather, there is a definite connection between their corresponding values such that variation in one variable (x) will cause variation in the other variable (y). As a functional relation, therefore, the above equation shows the dependence of one quantity (y) upon the other (x), y is called the dependent of variable (or the value of the function) and x the independent variable or (argument of the function).
Properties of a Function
Domain and Range: functions are studied in a set of real numbers R, which means an argument of a function can adopt only those real values at which a function is defined, i.e. it also adopts only real values. A set x of all admissible real values of an argument x, at which a function y = f (x) is defined, is called a domain of a function. While the range of these values are called independent variables.
A function is the set of all output values and these are called dependent variables, they are values that correspond to the second components of the ordered pairs it is associated with.
Suppose a cost function is C = 30 + 10Q Where Q is output and C is cost, now if Q = 50, then C = 30 + 10(50) and C = 530. The range of dependent variable is 30
Types of functions: there are several types of functions which includes;
Linear function: This type of function is expressed in first degree, the highest power of the unknown variable is unity and it is in form of y = a + bx where a is the intercept and b is the slope. If a linear function is equated to zero then it forms a linear equation. The graph of a linear function is always a Straight line. The function y = 2x + 3 is a typical example of a linear function since the power of the unknown variable x is unity.
Quadratic function: This is a function with the highest power of oits unknown variable to be equals to 2. Quadratic function exists in form of y = ax2 + bx + c, where a b and c are constants and a≠ 0. When a quadratic function is equated to zero it forms a quadratic function (ax2 + bx + c = 0). A quadratic function when plotted on a graph gives a curve because it is of the second degree. An example of quadratic function is y = 4x2 + 2x +2
Polynomial Function: A polynomial function is a function of degree n where the power of the unknown variable can take the form of any real number n, it is expressed in form of y = a0xn + a1xn-1 + ------ + an-1x + an (where a0 ≠ 0 and ai are constants, n is a non-negative integer x is a variable). Example of polynomial function of degree 4 is y = 2x4 + 3x3 +5x2 + x – 100.
Logarithmic and Exponential functions: Logarithmic functions existed in the form of x = Loga N, that is x is a logarithmic of N to the base of a so that ax = N. Example of logarithmic function is Log3 27 = 3. Exponential function on the other hand is expressed in form of y = ep where is the value of the natural exponential function (e = 2.71828) and p is a polynomial of degree n. Example of an exponential function is y = e5x3 + 2 . Logarithmic functions are the inverse of exponential functions. For example, the inverse of y = ax is y = logax, which is the same as x = ay. This definition is explained by knowing how to convert exponential equations to logarithmic form, and logarithmic equations to exponential form. The expression below demonstrates how to convert from exponential to logarithmic form and from logarithmic to exponential form.
Equations containing logarithms can be solved by changing everything in logarithmic form to exponential form. For example log2 x = -3 can be converted to exponential form as; 2-3 = x, and x = (1/8)
Evaluation of Function
When a function is represented by an equation, functional notation, f(x) is used. Consider the function
y = f(x) = 3x – 5
When a function is evaluated, a value is substituted into the function, to evaluate this function at f = 2 and f = (x+1) we have
f(x) = 3x – 5
f(2) = 3(2) – 5 = 6 – 5 = 1
f(x+1) = 3(x+1) – 5 = 3x + 3 – 5 = 3x – 2
Example 1
Find the values of the following functions;
y = f (x) = 6x3 + 7x2 – 20
When (i) x= 4
(ii) x = -3
Solution
(i) y = f (x) = 6x3 + 7x2 – 20
= f (4) = 6(4)3 + 7(4)2 – 20
= 6(64) + 7(16) – 20
= 384 + 112 –20
f(4)= 476.
(ii)y = f (-3) = 6(-3)3 + 7 (-3)2 – 20
= 6(-27) + 7(9) – 20
= -162 + 63 –20
f(-3) = -119.
Example 2
Given the function f(h) = 8h2 + 2h – h + 5
Find (i) f (-2) (ii) f (10) (iii) f (6)
(i) f (h) = 8h2 + 2h – h + 5
f (-2) =8(-2)2 +2 (-2) – (-2) + 5
=8 (4) – 4 + 2 + 5
= 32 – 4 + 10
f (-2)= 38.
(ii). f (10) =8(10)2 +2 (10) – 10 + 5
=8 (100) +20- 10 + 5
= 800 + 25 - 10
f(10)= 815.
(iii) f (6) =8(6)2 +2 (6) – 6 + 5
=8 (36) + 12 – 6 + 5
= 288 + 12 – 6 +5
f(6)= 299.
Example 3
If t(x) = 1/(√x2 – 16)
Find the domain of t and t(-6)
Solution
The domain of t is the set S = {x: x2 > 16}
And t(-6) = 1/ √ (-6)2 – 16
= 1/ √ 36 – 16
= 1/ √ 20
= 1/ 4.47
t (-6) = 0.22
Graphs
A graph shows the relationship between two variables in an equation. Graph is usually drawn using a grid system known as Cartesian (rectangular) coordinate system. The Cartesian is named after its developer, René Descartes (1596-1650). There are two intersecting axis that form four quadrants in a graph. The horizontal axis is called the x-axis. The vertical axis is called the y-axis. In a graph the point of intersection of the two axes is called the origin. The coordinates or the value of the x and the value of the y determine the point. This is also called an ordered pair (x, y). If a function is plotted on a graph, the point where it passes through the y axis is called y-intercept, also where it passes the axis is called the x- intercept.
Example 4
If y = 3x – 4 Find the y intercept, the x intercept and the slope
Solution
Slope is 3
y-intercept is where x = 0 and that is (0, -4)
x intercept is where y = 0
0 = 3x – 4
3x = 4
x = 4/3
x- Intercept is (4/3, 0)
Example 5
Write the equation If y = Ix + 3, in form of Slope – intercept form
Solution
Slope is I
y-intercept is where x = 0 and that is (0, 3)
x-intercept is where y = 0 and that is (-2/3, 0)
Example 6
Write the equation 6x = 8y + 7 in slope-intercept form
Solution
Solve for y
6x = 8y + 7
– 8y = – 6x + 7
y = Ox – Q
y = Hx – Q
slope is H
y-intercept is where x = 0 and that is (0, -Q)
x-intercept is where y = 0 and that is (7/6, 0)
Example 7
Plot the graph of y = x2 +10 where -2 < x < 3
Solution
X | -2 | -1 | 0 | 1 | 2 | 3 |
Y | 14 | 11 | 10 | 11 | 14 | 19 |
Example 8
Plot the graph of y = -2x2 for the value of x from – 5 to + 4
Solution
x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
y | -50 | -32 | -18 | -8 | -2 | 0 | -2 | -8 | -18 | -32 |
Slopes
When graphed, lines slope from left to right. However, some slope upward and others slope downward. Some are really steep, while others have a gentle slope. The slope of a line is defined as the change in y over the change in x, or the rise over the run.
This can be explained with a formula: (y2 - y1)/(x2 - x1) where (x1 ≠ x2). To find the slope, any two points on the line can be picked, the change in y is found, and then divided by the change in x.
This can be explained with a formula: (y2 - y1)/(x2 - x1) where (x1 ≠ x2). To find the slope, any two points on the line can be picked, the change in y is found, and then divided by the change in x.
Example 9
If The points (1,2) and (3,6) are on a line. Find the slope of the line
Solution:
m = (y2 - y1)/(x2 - x1)
m = 6 - 2
3 - 1
m = 2
When the slope and a point on the line are known, we can use the point-slope form to determine the equation of the line.
Example 10
Find the equation of the line passing through the following coordinates (1, 4) (3, 2)
Solution
Slope m = (y2 - y1)/(x2 - x1)
= 2 – 4
3 – 1
m = – 1
Recall that the straight line is given by
y - y1 = m (x -x1 )
y - 4 = m (x -1)
y - 4 = mx - m (substitute m = -1)
y - 4 = -x + 1
y = -x + 5
Applications of Functions to Social Sciences
In social sciences a lot of functional relationships exist between two or more variables. For example we have cost function, revenue function, profit function and consumption function. Demand and supply functions of a product are needed in the determination of equilibrium price and quantity respectively. The demand function is the relationship between quantity demanded of a commodity and its price and that represent the level of purchase of that particular product. If p is the price of the product and qd is the quantity demanded, then the demand function is given by the expression qd = f(p) and the graphical representation of the demand function gives the demand curve. If on the other hand p and qs represent the price and quantity supplied of a commodity, then, the supply functions is expressed as qs = f(p) and as in the case of demand the graphical representation of the supply function gives the supply curve. The intersection of the demand and the supply curves gives the equilibrium price and quantity respectively.
Example 11
If the demand and supply fuctions of a product are given by the functions qd = 20 – 4p and qs = 14 + 2p respectively. Calculate the equilibrium price and quantity for the product.
Solution
At Equilibrium, qs = qd
14 + 2p = 20 – 4p
2p + 4p = 20 - 14
6p = 6, and p = 1
Recall the demand Function;
qd = 20 – 4p and Substitute p = 1
= 20 – 4(1)
qd = 16
The equilibrium price is 1 and the equilibrium quantity is 16
Another important area in which fuctions are applied is in the break-even analysis. Break-even analysis is a very useful and relatively simple tool for management to use in making decisions. It can be applied in dealing with unknown variables such as demand by specifying the levels of known variables such as costs or profits, a required or minimum level can be found for unknown variable.. Break-even analysis has served as a substitute for estimating an unknown factor in making projects decisions. In deciding whether to go ahead or to skip it, there are always variables to be considered which include; demand cost, price and other miscellaneous factors. Break-even analysis is also useful in sales, profits and costs analyses it can also be useful to the issue of employing idle plant capacity, planning, advertising, granting credits and expanding production. Break-even point can be determined either through by the use of graphs or by formula. The point can be expressed either in terms of quantity or Naira sales or as percentage of capacity. At breakeven total revenue is equals to total cost given by the formula
BE (Q) = FC___
SP - VC
Where Q = Units produced
SP = Selling price per unit
VC = Variable cost per unit
In terms of Naira Sales break-even point is given by;
BE (N) = BE (Q).SP
As percentage of capacity the break-even point is given by;
BE(% CAP) = BE (Q) .100
CAP
Where CAP = Production Capacity
Example 12
The Marketing Department of Gyanawa Manufacturing Company is introducing a new product
based on the following information;
Fixed cost N 6500,
Per unit variable cost is N 315
Selling price per unit is N 380
Production capacity per period is 650 units
Calculate the break-even point in terms of units, sales revenue and as a Percentage of capacity
respectively.
Solution
FC = N 6500, VC = N 315, SP = N 380 and CAP = 650
B.E (Q) = FC____
SP-VC
= ___6500____
380 – 315
= 6500
65
B.E (Q) = 100 Units
BE (N) = BE (Q).SP
= 100.380
= N 38,000
As percentage of capacity
% CAP = 100 (100)
650
= 15.4%
Progress Test Questions
1. The demand and supply of money are given by md = 400 – 66i and ms = 260 + 88i. Find the equilibrium rate of interest and the money supply ?
- A manufacturer observed that his fixed cost of production is
N20000, variable cost per unit isN700 and selling price per unit isN920. Find his break-even point in terms of units and in Naira?
3. The consumption function of a hypothetical economy is given by c = a + byd while the autonomous consumption is N250 million marginal propensity to consume is 65% tax is N 12 million, and income is N540 million. Find the total consumption in the economy?
4. Write the equation If y =- 6/5x + 1/5, in form of Slope – intercept form.
- Plot the graph of y = 10x2 + 2x - 2 for the value of x from – 3 to + 3
- Plot the Graph of y = 12x2 – 5x + 20, Taking the values of x from – 4 t0 + 4, from the graph find the roots of the Equation?
- If g(x) = 1/(√x2 – 100) Find the domain of g and g(2)
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