Assignment 13

course Mth 158

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

09:35:34

**** query 1.5.34 (was 1.5.24). How did you write the interval [0, 1) using an inequality with x? Describe your illustration using the number line.

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0 <= x < 1

On the number line you would have a bracket on the left and a ) on the right and shade the area between the [ and )

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09:35:46

** My notes here show the half-closed interval [0, 1).

When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. **

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09:36:26

**** query 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'?

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x < -4 then x + 4 < 0

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09:36:32

** if x<-4 then x cannot be -4 and x+4 < 0.

Algebraically, adding 4 to both sides of x < -4 gives us x + 4 < 0. **

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09:37:09

**** query 1.5.46 (was 1.5.36). How did you fill in the blank for 'if x > -2 then -4x ____ 8'?

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x > -2 then -4x < 8

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09:37:14

**if x> -2 then if we multiply both sides by -4 we get

-4x <8.

Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. **

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09:38:04

**** query 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.

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2x + 5 >= 1

2x >= -4

x > -2

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09:38:10

** Starting with

2x+5>= 1 we add -5 to both sides to get

2x>= -4, the divide both sides by 2 to get the solution

x >= -2. **

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09:39:58

**** query 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.

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8 - 4(2-x) <= 2x

8 - 8 + 4x <= 2x

4x <= 2x

2x <= 0

x <= 0

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09:40:02

** 8- 4(2-x)<= 2x. Using the distributive law:

8-8+4x<= 2x. Simplifying:

4x<=2x. Subtracting 2x from both sides:

2x<= 0

x<=0 **

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09:47:19

**** query 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.

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0 < 1 - 1/3 x < 1 multply by -3

0 < 3 - 1x < 3

-3 < -1x < 0 divid by -1

3 > x > 0

0 < x < 3

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09:47:24

** Starting with

0<1- 1/3x<1 we can separate this into two inequalities, both of which must hold:

0< 1- 1/3x and 1- 1/3x < 1. Subtracting 1 from both sides we get

-1< -1/3x and -1/3x < 0. We solve these inequalitites separately:

-1 < -1/3 x can be multiplied by -3 to get 3 > x (multiplication by the negative reverses the direction of the inequality)

-1/3 x < 0 can be multiplied by -3 to get x > 0.

So our inequality can be written 3 > x > 0. This is not incorrect but we usually write such inequalities from left to right, as they would be seen on a number line. The same inequality is expressed as

0 < x < 3. **

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09:49:46

**** query 1.5.94 (was 1.5.84). Explain how you found a and b for the conditions 'if -3 < x < 3 then a < 1 - 2x < b.

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-3 < x < 3 then a < 1 - 2x < b

-3 < x < 3 Multiply by -2

6 > -2x > -6 add 1 to each

7 > -2x + 1 < -5

-5 < -2x + 1 < 7

a = -5

b = 7

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09:49:58

** Adding 1 to each expression gives us

1 + 6 > 1 - 2x > 1 - 6, which we simplify to get

7 > 1 - 2x > -5. Writing in the more traditional 'left-toright' order:

-5 < 1 - 2x < 7. **

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09:52:08

**** query 1.5.106 (was 1.5.96). Explain how you set up and solved an inequality for the problem. Include your inequality and the reasoning you used to develop the inequality. Problem (note that this statement is for instructor reference; the full statement was in your text) commision $25 + 40% of excess over owner cost; range is $70 to $300 over owner cost. What is range of commission on a sale?

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x = owners cost

.40 * 70 + 25 < .40x + 25 < .40 * 300 + 25

53 < .40x + 25 < 145

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09:52:16

** If x = owner cost then

70 < x < 300.

.40 * owner cost is then in the range

.40 * 70 < .40 x < .40 * 300 and $25 + 40% of owner cost is in the range

25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or

25 + 28 < 25 + .40 x < 25 + 120 or

53 < 25 + .40 x < 145. **

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09:54:52

**** query 1.5.112. Why does the inequality x^2 + 1 < -5 have no solution?

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x^2 + 1 < -5

x^2 < -6

x < sqrt(-6)

cant take the sqrt of a negative number

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09:55:21

STUDENT SOLUTION: x^2 +1 < -5

x^2 < -4

x < sqrt -4

can't take the sqrt of a negative number

INSTRUCTOR COMMENT: Good.

Alternative: As soon as you got to the step x^2 < -4 you could have stated that there is no such x, since a square can't be negative. **

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Assignment 13

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