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Exercises of Mathematics for Business

Claudio Mattalia

October 2010

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ii

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Chapter 1

Inequalities

1.1 Definitions

An inequality is an expressione of the kind:

A(x) B(x) x R

Solutions of the inequality are those values of the unknown that satisfy thisexpression.

Two inequalities are said equivalent if they admit the same solutions; holdin this sense the two basic principles:

1. Adding or substracting to both members of the inequality the same (con-stant or variable) quantity, an inequality equivalent to the given one isobtained:

A(x)> B(x) A(x) +C(x)> B(x) +C(x)

2. Multiplying or dividing both members of an inequality for the same (con-stant) positive quantity, an inequality equivalent to the given one is ob-

tained, multiplying or dividing them for the same (constant) negativequantity an inequality equivalent to the given one is obtained by reversingthe direction of the inequality:

A(x)> B(x)

c A(x)> c B(x) if c >0

c A(x)< c B(x) if c

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4 CHAPTER 1. INEQUALITIES

Example 1 Solve the inequality:

x 5 0

Adding the (constant) quantity+5 to both members we get (appling the1st

principle of equivalence):

x 5 + 5 0 + 5

i.e.:

x

5

this is the solution.

Example 2 Solve the inequality:

2x+ 3 > x

Adding the (variable) quantityxto both members we get (appling the1stprinciple of equivalence):

2x+ 3 x > x x

i.e:

x+ 3 > 0

and then adding the (constant) quantity

3 to both members we get (stillapplying the 1st principle of equivalence):

x+ 3 3> 0 3

i.e.:

x >3

that is the solution.

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1.1. DEFINITIONS 5

Example 3 Solve the inequality:

4x >5

Multiplying both members for the (constant and positive) quantity1

4we get

(applying the2nd principle of equivalence):

1

4 4x > 1

4 5

i.e.:

x >5

4that is the solution.

Example 4 Solve the inequality:

4x >5

Multiplying both members for the same (constant and negative) quantity 14

we get (applying the2

nd

principle of equivalence and reversing the inequality):1

4

(4x)0

(ifa 0).The solution is then easily given by:

x ba

Example 5 Solve the inequality:

4x 12> 0

In this case the inequality is already written in the canonic form, applyingthe equivalence principles seen above we get:

4x >12

and then:

x >3

that is the solution.

Example 6 Solve the inequality:

4x 12> 0

In this case the inequality is not written in the canonic form (as a < 0),multiplying both members for1(and reversing the inequality) we get first ofall:

4x+ 12< 0

that is the inequality written in the canonic form (as a > 0). We then easilyget (applying the equivalence principles):

4x

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1.3. RATIONAL INTEGER INEQUALITIES OF 2ND DEGREE 7

1.3 Rational integer inequalities of 2nd degree

The rational integer inequalities of 2nd degree can always be reduced to thecanonic form:

ax2 +bx+c 0 with a >0

(ifa 0).To solve an inequality of this type, first of all it is necessary to consider theassociated2nd degree equation:

ax2 +bx+c= 0

and to compute its roots x1 andx2 (where it is assumed x1 < x2 in the case ofdistinct roots).

We now have the following rule:

For values ofx external to the interval having as extremes the roots of theequation (that is for x < x1 and for x > x2) the trinomial ax2 +bx+chas the same sign of the coefficient a.

For values ofx internal to the interval having as extremes the roots of the

equation (that is for x1 < x < x2) the trinomial ax2 +bx + c has theopposite sign of the coefficient a.

For values ofx equal to the roots of the equation (that is for x= x1 andfor x = x2) the trinomial ax

2 +bx+c is null.

More precisely, keeping in mind that an inequality of2nd degree can have2real distinct roots, 2 real coincident roots or no real roots, the following threecases can be distinguished:

1. =b2 4ac >0. In this case the associated equationax2 +bx+c = 0has2 dinstict real roots x1, x2 (where it is assumed thatx1 < x2) and forthe trinomial ax2 +bx+c we have:

ax2 +bx+c >0 for x < x1 x > x2

ax2 +bx+c

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8 CHAPTER 1. INEQUALITIES

2. =b2 4ac= 0. In this case the associated equation ax2 +bx+c = 0has 2 real coincident roots x1 =x2 and for the trinomil ax

2

+bx+c wehave:ax2 +bx+c >0 for x =x1, x2

ax2 +bx+c= 0 for x= x1, x2

3. =b2 4ac 0 xR

The rule stated above then turns out to be valid in general, keeping in mindthat if = 0 all the values of x different from the roots x1 = x2 are to be

considered external to the interval having as extremes the same roots (and thenthere are no values ofxinternal to such interval), while if< 0 all the valuesofx are to be considered external to the interval having as extremes the roots(interval that in reality turns out to be empty, since these roots do not exist, sothat also in this case there are no values ofx internal to it).

Example 7 Solve the inequality:

x2 2x 8> 0

In this case the inequality is already written in the canonic form, and the

associated equation:x2 2x 8 = 0

has two distinct real roots x1 =2and x2 = 4. Applying the rule seen beforewe have that the solution of the inequality is given by:

x 4

Example 8 Solve the inequality:

x2 + 2x+ 8> 0

In this case the inequality is not written in the canonic form, hence multi-pliying both members for1(and reversing the inequality) we get:

x2 2x 8< 0

whose associated equation (the same of the previous excercise) has roots x1 =2and x2= 4. Applying the rule seen before we have that the solution of theinequality is given by:

2< x < 4

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1.3. RATIONAL INTEGER INEQUALITIES OF 2ND DEGREE 9

Example 9 Solve the inequality:

6x2 + 36x 0

whose associated equation:

6x2

36x

= 0

has roots x1= 0andx2= 6. The solution of the inequality is then given by:

x 6

Example 10 Solve the inequality:

x2 9

First of all the inequality can be written in the canonic form:

x2 9 0

then it is possible to observe that the associated equation:

x2 9 = 0

has rootsx1=3and x2 = 3,so that the solution of the inequality is given by:

x 3 x 3

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10 CHAPTER 1. INEQUALITIES

Example 11 Solve the inequality:

x2 2x+ 1 < 0

In this case the associated equation:

x2 2x+ 1 = 0

has two real coincident roots x1 = x2 = 1, and applying the rule seen beforewe have that the inequality is never satisfied (the same result can be obtainedobserving that the first member of the inequality is simply (x 1)2 that, beinga square, can never be 0

In this case the associated equation:

6x2 + 5 = 0

has no real roots, applying the rule seen before we then have that the inequalityis satisfiedx R (the same result can be obtained observing that the firstmember of the inequality is the sum of a non-negative term, 6x2, and of apositive term, 5, therefore it is strictly positive for every value of x, and theinequality is always satisfied).

1.4 Rational fractional inequalities

The rational fractional inequalities are those in which the unknown appears in

the denominator of a fraction, and they can always be reduced to the canonicform:

N(x)

D(x) 0

In this case first of all it is necessary to exclude the values ofx that makethe denominator of the fraction D(x) equal to 0 (since a fraction with nulldenominator has no meaning), then it is possible to study separately the signofN(x) and that ofD(x) and, combining them through the rule of signs, todetermine the sign of the fraction and to solve the inequality.

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1.4. RATIONAL FRACTIONAL INEQUALITIES 11

Example 13 Solve the inequality:

x+ 1

4x 8 >0

First of all it must be 4x 8= 0, from which x= 2 (condition of realityof the fraction). Studing separately the sign of the numerator and that of thedenominator of the fraction we then have:

N(x)> 0 x+ 1 > 0 x >1

D(x)> 0 4x 8> 0 x > 2

The sign of N(x) and of D(x), together with that of the fraction, can be

represented graphically in the following way (where the continous line indicatesthe intervals in which the sign is positive and the dashed line the intervals inwhich the sign is negative, while the cross indicates the value excluded from theexistence range):

From the analysis of this graphic we have that the solution of the inequalityis given by:

x 2

Example 14 Solve the inequality:

x+ 4x2 10

First of all it must be x2 1= 0, from which x=1 (condition of realityof the fraction). Studying the sign of the numerator and of the denominator ofthe fraction we then have:

N(x) 0 x+ 40 x 4

D(x)> 0 x2 1> 0 x < 1 x >1

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12 CHAPTER 1. INEQUALITIES

and graphically:

and the