Coordenadas, Gráficos e Retas

8/12/2019 Coordenadas, Grficos e Retas

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100 Chapter 1 Coordinates, Graphs, Lines

we only allowx >1. Thusx >2 is cut back tox >1

This is the set of solutions for our inequality in Case 1.

Case 2:

x+ 1


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Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 101

x+ 1, and indicate by+andsigns where each term is positive or negative. Thendraw a third line for (x+2)/(x+1); the signs along this line are determined by

multiplying the signs of the previous two lines.

For example, above3 we findand, the product of which is+. Hence weassign + to 3. On the other hand, above 3

2we find + and , the product of which

is . Hence we assign to 3

2 . Determining the signs in this way can be done quicklyand the final result shows that(x+ 2)/(x+ 1)is positive on the set

(, 2) (1, +)This agrees with the answer in our previous solution.

Example B

Solve(5 x)(x

+2)


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From this we can see that 2x3x+2 is positive wheneverx 32 . Hence the

solution set for our original inequality is

(, 2)

32

, +

Example C

Solve 1(x+ 1) 0, then thecorresponding interval is in the solution set for f(x) > 0. Dont let the length

of this discussion mislead you: the interval method is an easy-to-use technique.

Here are two examples, the first of which is a reworking of Example C:

Example D

Solve1

(x+ 1) 0?

(

,

3)

4

1

15

NO, since fx0

(1, 1) 0 3 NO, since fx

0

(Notice how we chose thex-values so that each f(x)is easy to compute.) Thusthe solution set is (

3,

1)

(1,

+), which agrees with the answer arrived at in

Example C.

Example E

Solve 0

x 24 x+ 8


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Solution

Since the expression

f(x)

=

x 2

4 x+ 8contains the terms

xand

x+ 8,

we must havex0 andx+ 80 (since we cannot take the square root of negativenumbers). Hence we are restricted to considering only thosex-values for whichx0. (These are thex-values for whichbothx0 andx 8 are true.)Step 1 Crossing points. We can have

x 2

4 x+ 8 =0

only when the numerator equals zero, i.e., when

x 2 = 0x = 2x = 4,

the only crossing point.

Step 2 Missing points. Having already restricted our attention tox 0, we willnot have any negative numbers occurring under square root signs (which would

make the expression undefined). Thus our expression is undefined only when the

denominator is zero, i.e., when

4

x+ 8 = 0x

+8

= 4

x+ 8 = 16x = 8,

the only missing point.

Step 3 The pointsx=4 and 8 divide the positivex-axis into three open intervals:


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Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 107

Is interval in solution set

interval x f

x

for f (x)0?

(0, 4) 1 1 NO, since fx

0

(8, +) 17 2

17 NO, since fx


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Notice that if (a) holds true, then (b) must be true. However, if (b) is true, then it

does notfollow that (a) is true. For example,x= 2 is in the solution set of (b), but itisnotin the solution set of (a). Thus (a) implies (b) isnota reversible step because

it isnottrue that (b) implies (a).

Example F

Solve

3x 2


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Section 1.2 Absolute Value 109

If the steps used to show that a statementAimplies a statementBare all reversible,

then we can say that Ais true if and only if Bis true. The term if and only if is

commonly abbreviated toiff.

1.2 ABSOLUTE VALUE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.2.1 The Definition of Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

When asked to find the absolute value of a specificnumberfor example, 5,23,

the chance that youll make a mistake is very small:

|5|=5,2

3

= 2

3, | |=

Perhaps it is because theintuitive ideaof absolute value is so simple that many peoplenever bother to learn theprecisedefinition:

|a|=

aifa0aifa


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Case 2: x


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Common Errors:

|a+ b| isNOTequal to|a| + |b||a b| isNOTequal to|a| |b|

There are, in fact, no simple rules which equate |a+ b| or|a b| with expressionsinvolving only |a| and |b|. The best we can do with either |a+ b| or|a b| is to boundthem by inequalities as follows:

||a| |b|||a+ b||a| + |b|||a| |b|||a b||a| + |b|

Of the four inequalities shown here, by far themost important is theupper bound

for

|a

+b

|:

|a+ b||a| + |b|Known as the triangle inequality, it is proved in Antons Theorem B.5. The other

three inequalities can be derived from clever applications of the triangle inequality.

This is the subject of Antons Exercises 4143in Appendix B.

Here is another common (and oftentimes costly) mistake involving absolute val-

ues:

Common Error:

a2 isNOTalways equal toa.The correct result uses absolute values (Theorem B.2):

a2 =|a| (1.1)

This occurs because the symbol

xdenotes thenon-negativesquare root ofx , e.g.,

(3)2 =

9=3=|3|Carefully read over Antons discussion of this preceding Theorem B.2.

Example C

Simplify the expression 1

x2 2x+ 1.


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Solution

1

x2 2x+ 1 = 1

(x 1)2= 1 |x 1| by (1.1)=

1 (x 1)=2 x ifx11 + (x 1)=x ifx


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andbare positive or negative, and regardless of whethera


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Thus (, 4) 1

3, +

is the solution set for the 1st inequality.

Second inequality:

0


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Step 1 Crossing points.

|x+ 4| |2x 3| = 0|x+ 4| = |2x 3|

x+ 4 = 2x 3 orx+ 4= (2x 3)x = 7 orx= 1

3

(Here we have used solution techniques similar to those employed in An-

tons Example 2in Appendix B.)

Step 2 Missing points. There are none since|x+ 4| |2x 3|is definedfor all values ofx .

Step 3 The pointsx

= 13and 7 give us three intervals:

Is interval in

solution set

interval x fx

for f (x)>0?, 1

3

1 2 NO

1

3, 7

0 1 YES

(7, +) 10 9 NO

Thus the solution set is1

3, 7

, agreeing with the first solution.

1.2.5 Another Absolute Value Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Inequalities of the form

k

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