course Mth 158
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
022. `* 22
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Question: * 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?
{(-2,5) (-1,3) (3,7) (4,12)}
The domain of the given numbers are:
(-2, -1, 3, 4) and the range of the given numbers are:
(5, 3, 7, 12)
Since the domain corresponds to exactly one element in the range, then it is a function.
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Given Solution:
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,-1,3,4)
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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Self-critique (if necessary):
There are no discrepancies, but the given solution has an error in the domain. The domain is (-2, -1, 3, 4)
Thanks for pointing that out. I've now corrected the typo in the original document as well as here.
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Self-critique Rating: 3
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Question: * 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)?* 3.1.30. y = (3x-1)/(x+2)
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Your solution:
This is somewhat confusing to look at, so I copied the problem from the book like this: (I did notice that the question here and the one in the book are slightly different. This one has (x+1)^2 and the book has (x+2)^2
Find the following for each function
46. f(x) = 1-1/(x+2)^2
(a) f(0) 1 – 1/(x+2)^2
1 – 1/(0+2)^2
1 – ¼
¾
(b) f(1) 1-1/(x+2)^2
1-1/(1+2)^2
1-1/9
8/9
(c) f(-1) 1-1/(x+2)^2
1-1/(-1+2)^2
1-1/1
0
(d) f(-x) 1-1/(-x+2)^2
1-1/x^2-4x+4
(e) –f(x) (-1) – 1/(x+2)^2
(-1) – 1/x^2+4x+4
(f) f(x+1) 1 – 1/((x+1) + 2)^2
1 – 1/x^2+2x+5
(g) f(2x) 1-1/(2x+2)^2
1-1/4x^2 + 8x + 4
(h) f(x+h) 1-1/((x+h) + 2)^2
confidence rating:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
* * 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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Self-critique (if necessary):
I think (e) was the only discrepancy in the problems stated above. I don’t see how it because (1/(x^2 +4x +4)) -1
Is that just written different than mine? I had almost the same answer.
There was a typo in the last line of that answer. I've corrected it. Thanks again for pointing out an error.
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Self-critique Rating: 3
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Question: * 3.1.36 / 44 (was 3.1.30)
Is y = (3x-1)/(x+2) the equation of a function?
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Your solution:
Yes if x is a real number because the solution is only one solution and does not equal 2 outputs.
confidence rating: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
** 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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Self-critique (if necessary):
I didn’t mention that -2 could not be equal to x because you cannot divide by 0.
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Self-critique Rating: 3
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Question: * 3.1.54 (was 3.1.40). G(x) = (x+4)/(x^3-4x)
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Your solution:
In finding the domain of the function:
G(x) = (x+4)/(x^3-4x)
Since the function is telling us to add 4 to x and then divide that number by x cubed and subtract x multiplied by 4, we have to see what x cannot equal.
x cannot = 0
x cannot = 2
x cannot = -2
So the answer is x such that x does not equal 0, 2, -2
confidence rating: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
* * 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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Self-critique (if necessary):
Well, I went about it the long way and plugged in the numbers until I found what would make the denominator 0. I still have trouble factoring. I don’t see how factoring x out of the denominator helped to come up with the solution. Wouldn’t you still have to guess at x?
Once you've factored out x, the other factor is x^2 - 4. This is the difference of two squares and factors accordingly as (x-2) (x+2).
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Self-critique Rating: 3
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Question: * 3.2.12 (was 3.1.50). Pos incr exp fn 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?
In the vertical line test, it states that a graph is a function if every vertical line intersects the graph at one point. In this graph it does intersect the line at point (0, 1) so it is a function.
I’m not sure how to answer the domain of the graph. There are no given points other than the y intercept.
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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Self-critique (if necessary):
I see now that the domain consists of all real numbers.
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Self-critique Rating: 3
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Question: * 3.2.16 (was 3.1.54) Circle rad 2 about origin.
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Your solution:
This graph is a circle and intercepts at 4 places, so it is not a function.
confidence rating: 3
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Given Solution:
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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Self-critique (if necessary):
I made the mistake by including the x intercepts, when the vertical line test is only the y intercepts and it has only 2 points.
You don't necessarily have to use the y intercepts. Any x value between -2 and 2, not including -2 or 2, defines a vertical line which intecepts the circle at two points.
That is, for -2 < x < 2, the vertical line through (x, 0) passes through the circle at two points.
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Self-critique Rating: 3
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Question: * 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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Your solution:
Yes, it is a function. The vertical line intercepts at one given point (1/5, 5). The domain is all real numbers. The range is all possible y values.
confidence rating: 3
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Given Solution:
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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Self-critique (if necessary):
There were no discrepancies.
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Self-critique Rating: 3
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Question: * 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?
• If f(2)=1/2 what is value of B?
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Your solution:
f(0) = 2
(2 * 0 - B) / (3 * 0 + 4) = -B / 4
I am a little lost on this one. This one is a little more difficult than the others in the book.
confidence rating: 0
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Given Solution:
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 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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Self-critique (if necessary):
I tried to write it on paper and follow how it was solved, but it is still a little confusing to me. Especially the second part where f(2) = ½ I don’t see where you plug in the ½ or where you plug in the 2?
f(x) = (2x - B) / (3x + 4), so
f(2) = (2 * 2 - B) / (3 * 2 + 4) = (4 - B) / 10.
Since f(2) = (4 - B) / 10,
f(2) = 1/2 means
(4 - B) / 10 = 1/2.
We solve this equation for B, as in the given solution.
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Self-critique Rating: 0
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Question: * 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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Your solution:
H(1) = 20 – 13(1)^2
H = 7 meters high at 1
H = 20 – 13(1.1)^2
H = 427/100
H = 4.27 meters high at 1.1
H = 20 – 13(1.2)^2
H = 32/25
H = 1.28 meters high at 1.2
15m = 20 – 13(x)^2
13x^2 = 20-15
13x^2=5
x^2 = .38
x = +- sqrt ( .38)
You can calculate sqrt(.38), which is about .62. So your solution is consistent with the given solution.
The same is pretty much the case with the rest of your answers. Take the square roots, and you've got the solutions.
I doubt this is right, but I gave it a shot anyway
10m = 20 – 13(x)^2
10 – 20 = -13x^2
-10 = -13x^2
10/13 = x^2
x = +- sqrt ( .77)
I tried it by subtracting 20 from both sides this time instead of moving the variable to the left of the equation. I have no idea if either are right.
5m = 20 - 13(x)^2
5-20 = -13x^2
-15 = -13x^2
15/13 = x^2
x = +- sqrt(1.15)
0m = 20 – 13(x)^2
-20 = -13x^2
x = +- sqrt(1.54)
confidence rating: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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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Self-critique (if necessary):
I didn’t get them all right, but it was close. At least I am making some progress on word problems.
See my note. You pretty much did get them right, needing only to take the square roots.
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Self-critique Rating: 3
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&#Good responses. See my notes and let me know if you have questions. &#