Assignment 22

022. `query 22

College Algebra

07-31-2008

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19:12:45

query 3.1.20 (was 3.1.6) (-2,5),(-1,3),(3,7),(4,12)}Is the given relation a function? Why or why not? If so what are its domain and range?

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RESPONSE -->

yes, the domain does not contain any repeating numbers

Domain: {-2, 4}

Range:{3,12}

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19:13:08

This relation is a function because every first element is paired with just one second element--there are no distinct ordered pairs with the same first element.

the domain is ( -2,-2,3,4)

the range is ( 5,3,7,12)

Another way of saying that this is a function is that every element of the domain appears only once in the relation.

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19:31:58

query 3.1.46 / 34 (was 3.1.20) f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h) for 1 - 1 / (x+1)^2What are you expressions for f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h)?query 3.1.30. y = (3x-1)/(x+2)

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RESPONSE -->

1-1/(0+1)^2

0/1

1-1/(1+1)^2

0/4

1-1/(-1+1)^2

0/0

0

-1+1/-(x+1)^2

0/-(x^2+2x+1)

1-1/x+1+1)^2

0/(x^2+4x+4)

1-1/(2x+1)^2

0/(4x^2+4x+1)

1-1/(x+h+1)^2

0/(x^2+h^2+1)

y = (3(0)-1)/(0+2)

-1/2

y = (3(1)-1)/(1+2)

2/3

y = (3(-1)-1)/(-1+2)

-4/1

-4

y = (3(-x)-1)/(-x+2)

-3x-1)/(-x+2)

y = -(3x-1)/-(x+2)

(-3x+1)/(-x-2)

y = (3(x+1)-1)/((x+1)+2)

3x+2)/(x+3)

y = (3(2x)-1)/(2x+2)

6x-1)/(2x+2)

y = (3(x+h)-1)/(x+h+2)

3x+3h-1)/(x+h+2)

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19:33:37

STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: f(x) = 1- 1/(x+2)^2

f(0) = 1- 1/ (0+2)^2

f(0) = 1-1/4

f(0) = 3/4

f(1) = 1- 1/ (3)^2

f(1) = 1- 1/9

f(1) = 8/9

f(-1) = 1- 1/(-1+2)^2

f(-1)= 1-1

f(-1)= 0

f(-x)= 1- 1/(-x+2)^2

f(-x)= 1 -1/ (x^2-4x+4)

-f(x) = -(1- 1/(x+2)^2)

-f(x)= -(1 - 1/ (x^2+4x+4))

-f(x) = (1/(x^2 + 4x + 4)) - 1

** Your answer is right but you can leave it in factored form:

f(-x) = -(1 - 1/(x+2)^2)

= -1 + 1 / (x+2)^2. **

f(x+1) = 1- 1/((x+1) + 2)^2. This can be expanded as follows, but the expansion is not necessary at this point:

= 1- 1/ ((x+1)^2 +2(x+1) + 2(x+1)+4)

= 1- 1/ ((x+1)^2 +8x+8)

= 1- 1/ (x^2+2x+1+8x+8)

= 1- 1/(x^2 + 10x +9)

** Good algebra, and correct, but again no need to expand the square, though it is perfectly OK to do so. **

f(2x)= 1-1/(2x+2)^2

= 1- 1/(4x^2+8x+4)

** same comment **

f(x+h)= 1- 1/((x+h)+2)^2

= 1- 1/((x+h)^2 + 4(x+h) + 4)

= 1- 1/ (x^2 + 2xh + h^2+4x+4h+4)

** same comment **

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RESPONSE -->

i see my mistake, i grouped the 1-1 together to get 0 every time.

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19:36:37

query 3.1.36 / 44 (was 3.1.30)

Is y = (3x-1)/(x+2) the equation of a function?

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RESPONSE -->

yes

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19:36:55

** This is a function. Any value of x will give you one single value of y, and all real numbers x except -2 are in the domain. So for all x in the domain this is a function. **

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19:43:55

query 3.1.54 (was 3.1.40). G(x) = (x+4)/(x^3-4x)

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RESPONSE -->

G(x) = (x+4)/(x^3-4x)

(x^3-4x)=x+4

x^3-5x-4=0

there is no y intercept

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19:45:16

** Starting with

g(x) = (x+4) / (x^3-4x) we factor x out of the denominator to get

g(x)= (x+4) / (x (x^2-4)) then we factor x^2 - 4 to get

g(x) = (x+4) / (x(x-2)(x+2)).

The denominator is zero when x = 0, 2 or -2.

The domain is therefore all real numbers such that x does not equal {0,2,-2}. **

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RESPONSE -->

ok i see what you wanted now. I didn't break it down far enough.

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19:45:57

**** query 3.2.12 (was 3.1.50). Pos incr exp fnDoes the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function?

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RESPONSE -->

?????

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19:46:15

using the vertical line test we determine this is a function, since any given vertical line intersects the graph in exactly one point.

The function extends all the way to the right and to the left, and there are no breaks, so the domain consists of all real numbers.

The range consists of all possible y values. The function takes all y values greater than zero so the range is {y | y>0}, expressed in interval notation as (0, infinity).

The y intercept is (0,1); there is no x intercept but the negative x axis is an asymptote.

This graph has no symmetery.

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RESPONSE -->

??

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19:46:41

query 3.2.16 (was 3.1.54) Circle rad 2 about origin.

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RESPONSE -->

This is not a function, it doesn't pass the vertical line test.

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19:46:42

query 3.2.16 (was 3.1.54) Circle rad 2 about origin.

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19:46:51

Using the vertical line test we see that every vertical line lying between x = -2 and x = 2, not inclusive of x = -2 and x =2, intersects the graph in two points. So this is not a function

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19:47:07

query 3.2.22 (was 3.1.60). Downward hyperbola vertex (1,2 5).Does the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function?

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RESPONSE -->

???

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19:47:14

Every vertical line intersects the graph at exacty one point so the graph depicts a function.

The function extends to the right and to the left without breaks so the domain consists of all real numbers.

The range consists of all possible y values.

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19:48:21

query 3.1.84 / 82 (was 3.1.70). f(x) =(2x - B) / (3x + 4).

If f(0) = 2 then what is the value of B?

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RESPONSE -->

2=0-B)/(0+4)

2=B/4

8=B

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19:48:52

If f(0) = 2 then we have

2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that

B = -4 * 2 = -8.

if f(2)=1/2 what is value of B?

If f(2) = 1/2 then we have

1/2 = ((2*2)-B) / ((3*2)+4)

1/2 = (4-B) / 10

5 = 4-B

1=-B

B=-1

**

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19:58:38

query 3.1.94 / 90 (was 3.1.80). H(x) = 20 - 13 x^2 (falling rock on Jupiter)What are the heights of the rock at 1, 1.1, 1.2 seconds?When is the rock at each altitude: 15 m, 10 m, 5 m.When does the rock strike the ground?

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RESPONSE -->

H(x) = 20 - 13( 1^2

H(x)=7

H(x) = 20 - 13 (1.1)^2

=20-15.73

h(x)=4.27

H(x) = 20 - 13 (1.2)^2

=20-18.72

h(x)=1.28

15=20-13x^2

-5=-13x^2

5/13=x^2

sqrt of (5/13)=x

10=20-13x^2

-10=-13x^2

10/13=x^2

sqrt of 10/13=x

5=20-13x^2

-15=-13x^2

15/13=x^2

sqrt of 15/13=x

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19:59:38

GOOD STUDENT SOLUTION: The height at t = 1 is

H(1) = 20-13

H(1) = 7m

The height at t = 1.1 is

H(1.1)= 20-13(1.1)^2

= 20-13(1.21)

= 20-15.73

H(1.1)= 4.27m.

The height at t = 1.2 is

H(1.2)= 20 - 13*(1.2)^2

= 20- 13 *(1.44)

= 20-18.72

H(1.2) = 1.28m.

The rock is at altitude 15 m when H(x) = 15:

15=20-13x^2

-5=-13x^2

5/13= x^2

x= +- .62

.62sec.

The rock is at altitude 10 m when H(x) = 10:

10=20-13x^2

-10=-13x^2

10/13 = x^2

x= +-.88

.88sec.

The rock is at 5 meter heigh when H(x) = 5:

5=20-13x^2

-15 = -13x^2

15/13=x^2

x= +- 1.07

1.07sec.

To find when the rock strikes the ground let y = 0 and we get

0= 20-13x^2. Adding -20 to both sides we have

-20=-13x^2. Multiplying both sides by -1/13 we get

20/13=x^2. Taking the square root of both sides we obtain the approximate value of x:

x=+-1.24

We conclude that

x = 1.24sec.

when the rock strikes the ground **

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You could use a little more detail in self-critiques, but you're doing OK here.