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Course description

References

iv iv Chapter 1. Applications of Diﬀerentiation

1.1. Mean value theorems of diﬀerential calculus

1.2. Using diﬀerentials and derivatives

1.3. Extreme Values

iii

1

1

5

7

Course description

Application of diﬀerentiation. Taylor theorem. Mean Value theorem of diﬀerential calculus. Methods of integration. Applications of integration.

References

1. Calculus: A complete course by Robert A. Adams and Christopher

Essex.

2. Fundamental methods of mathematical economics by Alpha C. Chiang.

3. Schaum’s outline series: Introduction to mathematical economics by Edward T. Dowling

iv

CHAPTER 1

Applications of Diﬀerentiation

1.1. Mean value theorems of diﬀerential calculus

Theorem 1.1.1 (Mean Value Theorem).

Suppose that the function f is continuous on the closed ﬁnite interval [a, b] and that it is diﬀerentiable on the interval (a, b). Then ∃ a point c ∈ (a, b) such that f (b) − f (a)

= f (c) . b−a It means that the slope of the chord joining the points (a, f (a)) and (b, f (b)) is equal to the slope of the tangent line to te curve y = f (x) at the point

(c, f (c)) so that the two lines are parallel.

Fig 2.28

Example 1.1.1.

√

Verify the conclusion of the mean value theorem for f (x) = x on the interval [a, b], where a ≤ x ≤ b.

Solution. We are to show that ∃ c ∈ (a, b) such that f (b) − f (a)

= f (c) b−a 1

XEQ 201 so long as f is continuous on [a, b] and is diﬀerentiable on (a, b). Now

√

√

1

f (x) = 2√x , f (a) = a, f (b) = b.

√

√

√

√

1

b− a b− a

1

∴ √ =

= √

=√

√ .

√

√

√

b−a

2 c b+ a b− a b+ a

The above equality implies that a < b, we have

√

a=

a+

2

√

a

√

√

c=

√

2

<

b+

2

√ b+ a

2

√

a

√

so that c =

√

2

<

b+

2

√

b

√

2

b+ a

.

2

Since

2

=b

which implies that c ∈ (a, b).

Example 1.1.2.

Show that sin x < x for all x > 0.

Solution. If x > 2π, then sin x ≤ 1 < 2π < x. If 0 < x ≤ 2π, then by

MVT, ∃ c ∈ (0, 2π) such that sin x sin x − sin 0 d =

= [MVT on [0, x]] = sin x x x−0 dx = cos c < 1 x=c which implies that sin x < x in this case too.

increasing decreasing functions

Definition 1.1.1 (Increasing and decreasing functions). Suppose that the function f is deﬁned on an interval I and that x1 and x2 are two points in I.

(a) If f (x2 ) > f (x1 ) whenever x2 > x1 , we say that f is increasing on

I.

(b) If f (x2 ) < f (x1 ) whenever x2 > x1 , we say that f is decreasing on

I.

(c) If f (x2 ) ≥ f (x1 ) whenever x2 > x1 , we say that f is non-decreasing on I.

(d) If f (x2 ) ≤ f (x1 ) whenever x2 > x1 , we say that f is non-increasing on I.

Diagram Fig 2.31

2

XEQ 201

Theorem 1.1.2.

Let J be an open interval and let I be an interval consisting of all points in J and possibly one or both of the end points of J. Suppose that f is continuous on I and diﬀerentiable on J.

(a)

(b)

(c)

(d)

If

If

If

If

f f f f (x) > 0

(x) < 0

(x) ≥ 0

(x) ≤ 0

for for for for all all all all x ∈ J, x ∈ J, x ∈ J, x ∈ J,

then then then then f f f f is is is is increasing on I. decreasing on I. non-decreasing on I. non-increasing on I.

derivative of increasing and decreasing functions

Example 1.1.3.

On what intervals is the function f (x) = x3 − 12x + 1 increasing? On what intervals is it decreasing?

Solution. f (x) = 3x2 − 12 = 3 (x − 2) (x + 2). It follows that f (x) >

0 when x < −2 or x > 2 and f (x) < 0 when −2 < x < 2. Therefore f is increasing on the intervals (−∞, −2) and (2, ∞) and is decreasing on the interval (−2, 2).

Diagram Fig 2.32

Example 1.1.4.

Show that f (x) = x3 is increasing on any interval.

Solution. Let x1 , x2 be any two real numbers satisfying x1 < x2 . Since f (x) = 3x2 > 0 for all x = 0, we have that f (x1 ) < f (x2 ) if either x1 < x2 ≤ 0 or 0 ≤ x1 < x2 . If x1 < 0 < x2 , then f (x1 ) < 0 < f (x2 ). Thus f is increasing on every interval.

Theorem 1.1.3.

If f is continuous on an interval I and f (x) = 0 at every interior point of

I, then f (x) = C, a constant on I.

3

derivetive of constant function

XEQ 201

derivative at interior extreme point Theorem 1.1.4.

If f is deﬁned on an open interval (a, b) and achieves a maximum (or a minimum) at the point c ∈ (a, b), and if f (c) exists, then f (c) = 0.

Values of x where f (x) = 0 are called critical points of the function f .

Theorem 1.1.5 (Rolle’s Theorem).

Suppose that the function g is continuous on the closed ﬁnite interval [a, b] and if it is diﬀerentiable on the open interval (a, b). If g (a) = g (b), ∃ a point c ∈ (a, b) such that g (c) = 0.

Theorem 1.1.6 (The Generalized Mean Value Theorem).

If functions f and g are both continuous on [a, b] and diﬀerentiable on (a, b), and if g (x) = 0 for every x ∈ (a, b), then ∃ a number c ∈ (a, b) such that f (c) f (b) − f (a)

=

. g (b) − g (a) g (c)

Exercise 1.1.

1. Illustrate the MVT by ﬁnding any points in the open interval (a, b) where the tangent line is parallel to the chord joining (a, f (a)) and

(b, f (b)).

(a) f (x) = x2 on [a, b]; Ans: c = b+a .

2

2

(b) f (x) = x3 − 3x + 1 on [−2, 2]; Ans: c = ± √3 .

2. Show that tan x > x for 0 < x < π .

2

3. Find intervals of increase and decrease of the following functions

2

(a) f (x) = x3 − 4x + 1. Ans: Increasing on −∞, − √3 and

2

√ ,∞

3

2

2

; decreasing on − √3 , √3 .

2

(b) f (x) = x2 − 4 . Ans: Increasing on (−2, 0) and (2, ∞); decreasing on (−∞, −2) and (0, 2).

(c) f (x) = x3 (5 − x)2 . Ans: Increasing on (−∞, 3) and (5, ∞); decreasing on (3, 5).

(d) f (x) = x + sin x. Ans: Increasing on (−∞, ∞).

4

XEQ 201

1.2. Using diﬀerentials and derivatives

Suppose dx is regarded as a new independent variable called the diﬀerential of x we can deﬁne a new dependent variable dy, called the diﬀerentail of y as a function of x and dx by dy dx = f (x) dx. dx For example if y = x2 , then dy = 2xdx means the same thing as dy/dx = 2x.

If f (x) = 1/x, then df (x) = − 1/x2 dx.

If y is a function of x, y = f (x), then denoting a small change in x by dx instead of ∆x, the corresponding small change in y, ∆y is approximated by the diﬀerential dy, i.e. dy =

∆y ≈ dy = f (x) dx.

Diagram Fig 2.25

Example 1.2.1.

Without using a scientiﬁc calculator, determine by a pproximately how much the value of sin x increases as x increases from π/3 to (π/3) + 0.006. To 3 decimal places, what is the value of sin ((π/3) + 0.006)?

Solution. If y = sin x, then dx = 0.006. Therefore

dy dx = cos x. Now x =

π

3

≈ 1.0472 and

dy π 1 dx = cos xdx = cos

· 0.006 = (0.006) = 0.003. dx 3

2

This means that the change in the value of sin x is approximately 0.003. Now dy =

sin

π π + 0.006 ≈ sin

+ 0.003 = · · · = 0.869(3 d. p.).

3

3

5

XEQ 201

Suppose changes in x are measured with respect to the size of x, then

relative change in x =

dx dx and percentage change in x = 100 . x x

Diﬀerentials and point elasticity1

For a demand function Q = f (P ), the elasticity is deﬁned as relative change in Q

∆Q/Q

=

.

∆P/P relative change in P point elasticity

If the change in P is inﬁnitesimal, then the expressions ∆Q and ∆P reduce to the diﬀerentials dP and dQ. In that case the elasticity measure assumes the sense of point elasticity of the demand function which is denoted by εd .

Thus

dQ/Q dQ/dP εd ≡

=

. dP/P Q/P

The numerator in the right hand is the derivative (or marginal2) function of the demand function while the denominator is the average function of the demand function. Thus the point elasticity is a ratio of the two functions.

In general, for any given total function y = f (x), the point elasticity of y

w.r.t. x is dy/dx marginal function εyx =

=

. y/x average function

The absolute value of the point elasticity measure is used in deciding whether the function is elastic at a particular point. In the case of a demand function,

elastic

|εd | > 1

The demand is of unit elasticity if |εd | = 1

inelastic |εd | < 1. at a given point.

1Elasticity is the measure of how an economic variable responds to change in another

variable. An elastic variable is one which responds more than proportionally to changes in other variables. An inelastic variable is one which changes less than proportionally in response to changes in other variables.

2Marginal denotes the rate of change of a quantity with respect to a variable on which it depends. 6

XEQ 201

Example 1.2.2.

Find εd for the demand function Q = 100 − 2P . Determine the point elasticity at P = 25.

Solution.

dQ dP = −2 and εd =

Q

P

=

100−2P

.

P

Therefore

−2

P

=

.

(100 − P ) /P

P − 50

Thus

= · · · = −1.

εd

P =25

Therefore the demand is of unit elasticity when P = 25.

Exercise 1.2.

1. Use diﬀerentials to determine approximate change in the values of the given function as its argument changes from the given value to the given amount. What is the approximate value of the function after the change?

(a) y = 1/x as x increases from 2 to 2.01.

(b) h (t) = cos (πt/4) as t increases from 2 to 2 + (1/10π).

2. Find the approximate percentage changes in the given function that will result from an increase of 2% in the value of x.

(a) y = x2

(b) y = 1/x2

3. Given the consumption function C = a + bY (with a > 0; 0 < b <

1);

(a) Find its marginal function and its average function.

(b) Find the income elasticity of consumption εCY , and determine its sign, assuming Y > 0.

(c) Show that the consumption is inelastic at all positive income levels. 1.3. Extreme Values

Maximum and Minimum Values

Definition 1.3.1. Function f has an absolute maximum value f (x0 ) at x0 in its domain if f (x) ≤ f (x0 ) holds for every x in the domain of f .

7

absolute extreme values

XEQ 201

Similarly, f has an absolute minimum value f (x0 ) at x0 in its domain if f (x) ≥ f (x0 ) holds for every x in the domain of f .

Remark 1. extreme value is unique, can occur at several points existence of extreme value not guaranteed

existence of extreme values for closed ﬁnite intervals local extreme values 1. A function will have only one absolute maximum (or minimum) value if it exists. However the value can occur at many points. For example, f (x) = sin x has absolute maximum of 1 but it occurs at every point π + 2nπ, n ∈ Z.

2

2. A function need not have any extreme value. The function f (x) =

1

x becomes arbitrarily large as x approaches 0 from the right, and so has no ﬁnite absolute maximum value.

Theorem 1.3.1.

If the domain of the function f is a closed, ﬁnite interval or a union of ﬁnitely many such intervals, and if f is continuous on that domain, then f must have an absolute maximum value and absolute minimum value.

Definition 1.3.2. Function f has a local maximum value (loc. max.) f (x0 ) at the point x0 in its domain provided ∃ a number h > 0 such that f (x) ≤ f (x0 ) whenever x is in the domain of f and |x − x0 | < h.

Similarly, f has a local minimum value (loc. min.) f (x1 ) at the point x1 in its domain provided ∃ a number h > 0 such that f (x) ≥ f (x1 ) whenever x is in the domain of f and |x − x1 | < h.

Diagram Fig 4.17

From the above ﬁgure we see that local extreme values can occur at any of the following points.

8

XEQ 201 critical, singular or end points

(i) critical points of f ; points x ∈ D (f ) where f (x) = 0.

(ii) singular points of f ; points x ∈ D (f ) where f (x) is not deﬁned.

(iii) endpoints of the domain of f ; points that do belong to D (f ) but are not interior points of D (f ).

In the ﬁgure above, x1 , x3 , x4 are critical points, x2 and x5 are singular points and a and b are endpoints.

Theorem 1.3.2.

If the function f is deﬁned on an interval I and has a local maximum (or local minimum) value at the point x = x0 in I, then x0 must be either a critical point of f , a singular point of f , or and endpoint of I.

Example 1.3.1.

Find the maximum and minimum values of the function g (x) = x3 − 3x2 −

9x + 2 on the interval −2 ≤ x ≤ 2.

Solution. Since g is a polynomial it can never have a singular point.

For critical points we calculate g (x) = 3x2 − 6x − 9 = 3 (x + 1) (x − 3) = 0.

Thus x = −1 or x = 3. But x = 3 is not in the domain of g and so we ignore it. We then investigate the endpoints x = −2 and x = 2 and critical point x = −1. g (−2) = 0, g (−1) = 7, g (2) = −20

The maximum value of g on −2 ≤ x ≤ 2 is at the critical point x = −1, and the minimum value is at the endpoint x = 2.

Diagram Fig 4.19

9

exteme values occur at critical, singular or end points

XEQ 201

Example 1.3.2.

Find the maximum and minimum values of h (x) = 3x2/3 −2x on the interval

[−1, 1].

Solution. The derivative of h is h (x) = 2x−1/3 − 2.

Note that x−1/3 is not deﬁned at x = 0 in D (h), so x = 0 is a singular point of h. Also h (x) = 0 at x−1/3 = 1, that is at x = 1, which also happens to be an endpoint of the domain of h. We therefore examine the values of h at endpoints x = −1 and x = 1 and at the singular point x = 0. h (−1) = 5, h (0) = 0, h (1) = 1

The function h has a maximum value 5 at the endpoint x = −1 and a minimum value 0 at the singular point x = 0.

Diagram Fig 4.20

10

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...Price Theory Assignment 1 – Consumer Theory Click Link Below To Buy: http://hwaid.com/shop/price-theory-assignment-1-consumer-theory/ Answers are in Image Format Suppose that Sally’s preferences over baskets containing petrol (good x), and food (good y), are described by the utility function U (x, y) = xy + 100y. The marginal utilities for this function are, MUx = y and MUy = x + 100. Use Px to represent the price of petrol, Py to represent the price of food, and I to represent Sally’s income. Question 1: Find Sally’s petrol demand function, and Sally’s food demand function. (8 Marks) Question 2: From Sally’s perspective, is food a normal good, an inferior good, or neither normal nor inferior? Brieﬂy explain with reference to your answer to question 1. (2 Marks) Question 3: Suppose that the price of petrol is $2 per litre, the price of food is $5 per kilogram, and Sally’s income is $400. What quantities of food and petrol does Sally consume? What level of utility does Sally receive from this consumption basket? (3 Marks) Question 4: Suppose that, as in question 3, the price of petrol is $2 per litre, the price of food is $5 per kilogram, and Sally’s income is $400. Now suppose that the government is considering two alternative policies to improve Sally’s welfare. Policy 1: Place a $0.4 per litre subsidy on petrol, reducing the price of petrol to $1.6 per litre. Policy 2: Give Sally a voucher that can be used to purchase food (but not......

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...Student’s Name: Course Name: Instructor’s Name: Institution: Date: Consumer choice theory is a microeconomics branch that tries to relate preferences to both consumer demand curves and consumer expenditures. The theory analyses the way consumers maximize their need to consume which is measured by their preferences against the limited ways on their expenditure. Consumers do this by utility maximization subject to a constraint on their budget. Other times it gets referred to as the theory of consumer behavior. Through the study of this theory, researchers can explain why the consumers would buy more of the product when its price is less as compared to when its price is high. Another elaboration of the theory is that it shows the reason why the households spend their income as they always do (Haugtvedt, Herr, & Kardes, 2008). The greater assumption is that every consumer is rational and aims at maximizing their satisfaction. Some major theories explain the consumer behavior. First is the Cardinalist approach or the marginal utility theory and the second is the ordinalist approach or the analysis of the indifference curves. The former describes extra satisfaction a consumer derives after consuming an extra unit of a commodity while consumption of all other products remains unchanged. The law of diminishing marginal utility gives a thorough elaboration on why the demand curves always have a downward sloping nature. The latter shows the line of combinations (indifference......

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...Consumer Behavior Models in Tourism Analysis Study Muhannad M.A Abdallat, Ph.D. Assistant Professor Hesham El –Sayed El - Emam, Ph.D. Assistant Professor Department of Tourism and Hospitality, Faculty of Tourism and Archeology King Saud University ABSTRACT The theories of consumer decision-making process assume that the consumer’s purchase decision process consists of steps through which the buyer passes in purchasing a product or service. However, this might not be the case. Not every consumer passed through all these stages when making a decision to purchase and in fact, some of the stages can be skipped depending on the type of purchases. The reasons for the study of consumer’s helps firms and organizations improve their marketing strategies by understanding issues such as: • The psychology of how consumers think, feel, reason, and select between different alternatives (e.g., brands, products); • The psychology of how the consumer is influenced by his or her environment (e.g., culture, family, signs, media); • The behavior of consumers while shopping or making other marketing decisions; • Limitations in consumer knowledge or information processing abilities influence decisions and marketing outcome; • How consumers’ motivation and decision strategies differ between products, that differ in their level of importance or interest that they entail for the consumer; and • How marketers can adapt and improve their marketing campaigns and marketing......

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...that is goal directed in nature. Three experiments demonstrate that the primed behavior showed typical goal-directed qualities, including increased performance postdelay, decreased performance postprogress, and moderation by motivation. P eople see thousands of brand images in an average day. Given how ubiquitous brands have become in people’s everyday lives, it is important that research uncovers the ways in which brand exposure can affect behavior. Although brands are of signiﬁcant interest to consumer researchers, scant empirical work has addressed the potential behavioral consequences of brand exposure, inside or outside of the consumer decision-making context. And yet, given that consumers encounter many more brands than people in an average day, brands have surely become more psychologically meaningful than the existing empirical work would suggest. Our ﬁrst objective is to investigate whether behavioral priming effects translate from the social to consumer do- main. Can brand primes elicit effects on behavior in the same fashion as can person primes? Our second objective is to understand underlying mechanisms. If brand primes can shape behavior, what is the process by which they elicit their effects? On the one hand, because brands are thought to be linked to personality traits, they may elicit cognitively based behavioral effects, as do person representations. On the other hand, brands are symbols of aspirations, representing desired......

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