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Университет имени Сулеймана Демиреля

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	42	Solve the inequality			x4	3x3 + 2x2
EXAMPLE					x4	3x3 + 2x2
							< 0.
						x3 5x2	< 0.
			x4 3x3 + 2x2			x3 5x2
Solution			x4 3x3 + 2x2	< 0
Solution			x3 5x2	< 0
			x3 5x2

x2 (x 1)(x 2)

x2 (x 5) < 0 x = 0 (double root) x = 1, x = 2, x = 5

–¥

x4 – 3x3 + 2x2

–

– 5x2

So x (– , 0) (0, 1) (2, 5).

43 Solve the inequality

x 1

x+1

EXAMPLE

x 1

< 2.

Solution

x 1

x+1

2 < 0 ; x 0

and

x 1

(x 1)2 x(x+1) 2x(x 1) 0

x(x 1)

2x2 x+1

< 0

x(x 1)

2x2

x+1= 0

x = 1 or x = 1

x(x 1) = 0

x = 0, x =1

–¥

–1

–2x2 – x + 1

–

x(x – 1)

So x

(– , –1) 0,

(1, ).

Equations and Inequalities

289

1 2x

EXAMPLE

Solve the inequality

x3 +1.

x+1

Solution

1 2x

;

x 1

x+1

x2 x+1

x3 +1

x2 x+1 2(x+1) (1 2x)

(x+1)(x2 x+1)

x2 x 2	0
(x+1)(x2 x+1)

x2 x 2 = 0 ; x = 1 or x = 2

x + 1 = 0 ; x = –1 (double root) x2 – x + 1 = 0 ; no real solution


		x	–¥		–1	2	¥

	x2 – x – 2			–	–		+
(x + 1)(x2		– x + 1)		–	–		+
(x + 1)(x2		– x + 1)

So x (– , –1) (–1, 2].


	45 Solve the inequality	2x2 +3x 2	. (1 x2 )		0.
EXAMPLE
		(x2 +3x) .		x 2

Solution

The absolute value of a number is never negative.

|x| 0, x

Since |2x2 + 3x – 2| and |x – 2| are non-negative, just check their roots.

2x2 + 3x – 2 = 0 ; x = –2, x =	1
	2

x – 2 = 0 ; x = 2

–2 and 1 satisfy the inequality, so these values are in the solution set. However, 2 is not in

the solution set because it makes the denominator zero.

1 – x2 = 0 ; x = –1, x = 1

x2 + 3x = 0 ; x = 0, x = –3

–¥

–3

–1

|2x2 + 3x – 2|(1 – x2)

–

(x2 + 3x)|x – 2|

So x (– , –3) [–1, 0) {–2,

} [1, 2) (2, ).

290	Algebra 11

x2 6x 16

EXAMPLE

Find the domain of the function

y = 4

+ x 1 if y is a real number.

x2 12x+11

Solution

EXAMPLE 47

7 3x4 4x3 + x 1 is a real number for all values of x because the index is an odd number.

x2 6x 16

x2 12x+11 is a real number if

12x+11

Therefore, y is a real number if

x2 6x 16

x2 12x+11

x2 6x 16 = 0 ; x = 2, x = 8

x2 12x+11= 0 ; x =1, x =11

–¥

–2

x2 – 6x – 16

–

x2 –12x + 11

So x (– , –2] (1, 8] (11, ).

Solve the inequality 3

3x 1

x 3

2 x 1 < 8

3x 7

Solution

If ab < ac then b < c.

3x 1

x 3

3x 1

x 3

2 x 1 < 83x 7 ;

2 x 1

<(23 )3x 7

3x 1

3 x 3

3 x 1

< 2

3x 7

3x 1

< 3(x 3)

3(x 1)

3x 7

3x 1

3x 9

< 0

3x 3

3x 7

(3x 1)(3x 7) (3x 9)(3x 3)				< 0
	(3x 3)(3x 7)
	12x 20		< 0

	(3x 3)(3x 7)
12x 20 = 0 ; x = 5
		3
3x 3 = 0 ;		x =1
3x 7 = 0 ;		x = 7
		3

Equations and Inequalities

291

						x	–¥		1		5		7	¥
											5		7
											3		3

		12x – 20						–		+		–		+

	(3x – 3)(3x – 7)
	(3x – 3)(3x – 7)
So x ( , 1)			5	,	7
So x ( , 1)			3	,	3	.
			3		3

Check Yourself 3

Solve the inequalities.

1.	x2 + 5x – 6 > 0				2.	(x + 3)3(x – 1)2(x – 4) 0				3.	2		+	3		4
1.	x2 + 5x – 6 > 0				2.	(x + 3)3(x – 1)2(x – 4) 0				3.		2 x	+	2+ x	4	x2
												2 x		2+ x	4	x2
	2		2					x
	(x 2)(x +4)						1	x
4.	(x 2)(x +4)				5.		1	2
4.		2	4	< 0	5.			4x x
		x	4				2
Answers
1.	(– , –6) (1, )				2.	(– , –3] {1} [4, )				3.	(–2, 2)			[6, )
4.	(–2, –ñ2) (ñ2, 2)				5.	(– , 0] [ 1			, )
								2

‘Obvious’ is the most dangerous word in mathematics.

292	Algebra 11

EXERCISES 4.4

1. Determine the sign of each polynomial.

a. x2 – 5x + 4		b.	2x2 + x – 6
c.	2x2 – 3x + 4	d. –16x2 + 8x – 1
e.	–4x2 + 10x – 25	f.	12x2 + 4ñ3x + 1

2. Solve the inequalities.
a. x2 < 9x – 20	b. 4x – 7x2 > 0
c. x(6x + 7) 0	d. 2x – 6 < 3x2
e. 3x2+8x 16	f. x(x 1)+1<	4 x+	1
3		5	5

g. (2x – 1)2 > (x – 5)2

3.For which values of k does the equation x2 + 2(1 – k)x + 1 = 0 have

a. no real solution?

b. one double root?

c. two distinct real roots?

4.For which values of b does the equation

3x2 + (b + 1)x + 1 = 0 have two distinct real roots?

5.For which values of a does the equation ax2 + (a + 1)x + 2 = 0 have no real root?

6.The product of a number and four plus the number is less than 15. Find all possible integer values of the number.

7. Solve the inequalities.

a. x(x – 1)2 > 0

b. (2 – x)(3x + 1)(2x – 3) > 0

c. (3x – 2)(x – 3)3(x + 1)3(x + 2)4 < 0

d. (x 1)(3x 2) > 0 5 2x

e. (x+1)(x+2)(x+3) > 0 (2x 1)(x+4)(3 x)


f.	x4	+ x2 +1		< 0
	x2	4x 5

g.	3x 2		< 3

	2x 3

2x2 +18x 4

> 2

x2 +9x+8

x+1

x 2

2 x

1 2x

x3 + x2

x3 3x2

3x 2 x2

7x 4 3x2

(x2 +4x+4)(x 2)1001

< 0

(x 2)2004

x+ 4

n. |6x2 – 2x + 1| 1

o. x+2 < 3 2x 3

p.x2 3x+2 >1 x2 +3x+2

q.x22 3x 1 1 x + x+1

Equations and Inequalities

293

8. Solve the inequalities.



a. 1

1 +

< 0

(a2

+ a+1)x2 + 4

> 0

a4 x

a4 x2

3 x(x2 + x+3)

< 0

2 x+

2x2 5x+2

x 3

5x+6

9. Find the domain of each function.

a.	y =	2

		3 x2 49
b.	y=	1

			x 4

c.	y =		3x 6
			x+2

d. 4 (x4 5x3 +6x2 )(1 x2 )

10. Solve the inequality		12		3	>1.

	x2	+ 2x	x2	+ 2x 2


11. Solve the inequality	x4 + 3x3 + 4x2 8	< 0.
	x2


294	Algebra 10

EXAMPLE 48

Solution

We saw in Chapter 1 that a set of simultaneons equations to solve is called a system of equations. A system that of equations includes more than one inequality is called an inequality system.

To find the solution of a system, we solve each inequality separately and then find the intersection of the solutions.

	x2	7x 8 > 0
Solve the inequality system		.
	x2	4x+ 3 > 0

First find the zeros of each polynomial.

(1)x2 – 7x – 8 > 0

x2 – 7x – 8 = 0 ; x = –1, x = 8

(2)x2 – 4x + 3 > 0

x2 – 4x + 3 = 0 ; x = 1, x = 3

We need to find values so that both polynomials are greater than zero. Let us check the chart.

x –¥	–1	1	3	8	¥
(1)	+	–	–	–	+
(2)	+	+	–	+	+
system

EXAMPLE 49

Solution

We can see that both polynomials are greater than zero when x (– , –1) (8, ). This is the intersection of the solutions.

x2 + x 4 Solve the inequality system x <1.

x2 < 64

(1)		x2 + x 4	<1 ;	x2	+ x 4 x	< 0 ;
		x			x

	x2 – 4 = 0 ; x = 2 ; x = 0
(2)	x2 < 64 ;		x2 – 64 < 0
	x2 – 64 = 0 ; x = 8

x2 4 < 0 x

x –¥	–8	–2	0	2	8	¥
(1)	–	–	+	–	+	+
(2)	+	–	–	–	–	+
system

This time, both polynomials need to be less than zero, so x (–8, –2) (0, 2).

Equations and Inequalities

295

100x3

50 Solve the inequality system (x+9)(5x x2

EXAMPLE

18)

x2 18x+45

Solution (1) x5 100x3

x5 – 100x3 0 ;

x3(x2 – 100) = 0

x = 0,

x = 10

(2)

(x+9)(5x x2

18)

x2 18x+45

x + 9 = 0 ; x = –9

5x – x2 – 18 = 0 ; no solution

x2 – 18x + 45 = 0 ; x = 3, x = 15

–¥

–10

–9

(1)

–

(2)

–

system

So x [–10, –9] [10, 15).

EXAMPLE	51 Solve the inequality		2x 1 < x+ 2.
		2x 1 0	(1)
Solution		x+ 2 > 0	(2)
		2	(3)
	2x 1<(x+ 2)		(3)
	(1) 2x – 1 0		(2) x + 2 > 0		(3)	2x – 1 < x2 + 4x + 4
		2x – 1 = 0	x + 2 = 0			x2 + 2x + 5 > 0
		1	x = –2			x2 + 2x + 5 = 0
If f(x) g(x)		x = 2				since < 0, there is no real solution
If f(x) g(x)						1
						1
f (x) 0			x –¥	–2		2	¥
				–	–		+
then, g(x) 0			(1)	–	–		+
	(x)
f (x) g2	(x)
			(2)	–	+		+
			(2)	–	+		+
			(3)	+	+		+
			system
	So x [ 1 , ).
		2

296	Algebra 11

EXAMPLE 52 Solve the inequality x2 + x 2 > x.

Solution

If f(x) g(x) then

g(x)< 0

f (x) 0 or

g(x) 0

f (x) g2(x)

Case 1

Case 2

__________________

x < 0

(1)

x 0

(1)

+ x 2 0 (2)

+ x 2 > x2

(2)

(1)

x < 0

(1)

x 0

(2)

x2 + x – 2 = 0 ;

(2)

x – 2 > 0 ;

x1 = –2, x2 = 1

x – 2 = 0, x = 2

–¥

–2

(1)

–

Case 1

(2)

–

	system
	(1)	–	–	+	+	+
Case 2	(2)	–	–	–	–	+
	system

So x (– , –2] (2, ).

	53	Solve the inequality (x2 + x + 1)2 – 4(x2 + x + 1) + 3 < 0.
EXAMPLE		Solve the inequality (x2 + x + 1)2 – 4(x2 + x + 1) + 3 < 0.

Solution		For the inequality, we let t = x2 + x + 1. Then the original inequality becomes
		t2 – 4t + 3 < 0. First let us solve the inequality for t.
		t2 – 4t + 3 = 0 ; t = 1, t = 3
						3
			x	–¥	1	3	¥

			t2 – 4t + 3		+	–	+

So 1 < t < 3. Now solve for x. 1 < x2 + x + 1 < 3

0 < x2 + x < 2, which gives us the system of inequalities

x2	+ x > 0	(1)
		.
x2	+ x 2 < 0	.
x2	+ x 2 < 0	(2)

Equations and Inequalities

297

(1)

+ x > 0

x2 + x = 0 ; x = 0, x = –1

(2)

+ x – 2

< 0

+ x – 2

= 0 ;

x = –2,

x = 1

–¥

–2

–1

(1)

–

(2)

–

system

So x (–2, –1) (0, 1).

EXAMPLE 54 Solve the inequality |x2 – 3x + 2| 2x – x2.

Solution			Case 1
	________________________________________
		x2	3x+ 2 0	(1)

		x2	3x+ 2 2x x2
		x2	3x+ 2 2x x2	(2)

	(1)		x2 – 3x + 2 0
			x2 – 3x + 2 = 0 ; x = 1, x = 2
	(2)		x2 – 3x + 2 2x – x2		; 2x2 – 5x + 2 0
			2x2 – 5x + 2 = 0 ; x = 1 , x = 2
					2
			Case 2
	_______________________________________
		x2	3x+ 2 < 0		(1)

		(x2 3x+ 2) 2x x2
		(x2 3x+ 2) 2x x2			(2)

	(1)		x2 – 3x + 2 < 0
			x2 – 3x + 2 = 0 ; x = 1, x = 2
	(2)		–(x2 – 3x + 2) 2x – x2
			–x2 + 3x – 2 2x – x2
			x – 2 0
			x – 2 = 0 ; x = 2

298	Algebra 11

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