Query 13

#$&*

course Mth 163

10/27 630

013. `query 13

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Question: `qQuery last asst before test 1, problem 1.Give your solution to x ^ 3 / 17 = 58

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Your solution:

x^3 * 17 = 58

x^3 * 17*17 = 58*17

x^3 = 986

x^(3*1/3) = 986^(1/3)

x = 9.95

confidence rating #$&*: 3

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Given Solution:

`a** The solution to x^3 / 17 = 58 is found by first multiplying both sides by 17 to get

x^3 = 58 * 17

then taking the 1/3 power of both sides, obtaining

(x^3)^(1/3) = (58 * 17)^(1/3) or

x = 9.95, approx..

COMMON ERROR:

If you interpret the equation as x^(3/17) = 58 you will get solution x = 58^(17/3) = 9834643694. However this is not

the solution to the given equation

To interpret x ^ 3 / 17 you have to follow the order of operations. This means that x is first cubed (exponentiation

precedes multiplication or division) then divided by 17. If you introduce the grouping x^(3/17) you are changing the

meaning of the expression, causing 3 to be divided by 17 before exponentiation. **

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `qGive your solution to (3 x) ^ -2 = 19

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Your solution:

3x^-2 = 19

3x^(-2 * -1/2) = 19^(-1/2)

3x = .229

x = .0763

confidence rating #$&*: 3

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Given Solution:

`a** (3x)^-2 = 19 is solved by taking the -1/2 power of both sides, or the negative of the result:

((3x)^-2)^(-1/2)) = 19^(-1/2) gives us

3x = 19^(-1/2) so that

x = [ 19^(-1/2) ] / 3 = .0765 or -.0765. **

STUDENT QUESTION:

I'm not sure how I was supposed to arrive at two solutions.

INSTRUCTOR RESPONSE

If an even power of a number has a given value, so does that power of the negative of that number.

It's fairly easy to understand why. For example:

5^2 = 25 and (-5)^2 = 25

3^4 = 81 and (-3)^4 = 81

2^(-4) = 1/16 and (-2)^(-4) = 1/16.

So when solving an equation of the form (c)^n = d, where n is an even integer (whether positive or negative) and d is a positive quantity, there are two solutions: c = d^(1/n) and c = - d^(1/n).

(note that - d^(1/n) is the negative of d^(1/n), not (-d)^(1/n), which would not be defined).

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Self-critique (if necessary):

Do I need to give two solutions? Or is one sufficient????

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Self-critique rating: ok

@&

Any solution to an equation that can be reduced to the form

x^n = c ,

where n is an even power, will have two such solutions.

*@

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Question: `qGive your solution to 4 x ^ -.5 = 7

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Your solution:

4x^-.5 = 7

4x^(-.5*-1/.5) = 7^(-1/.5)

@&

You would have to raise both sides to the -1/.5 power. That would include the 4.

You would get

( 4 x ^ (-.5) ) ^ (1 / (-.5)) = 7^(-1 / (-.5) )

which would give you

4 ^ (1 / (-.5)) * (x^-.5)^ (1 / (-.5)) = 7^ (1 / (-.5))

*@

4x = 7^-2

4x = .0204

X = .0051

confidence rating #$&*: 3

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Given Solution:

`a** to solve the equation we first multiply both sides by 1/4 to get

x ^ -.5 = 7 / 4. Then we raise both sides to the -2 power:

(x^-.5)^-2 = (7/4)^-2 so

x = .327 approx **

STUDENT QUESTION

Why do we multiply by ¼?

INSTRUCTOR RESPONSE

4 is not part of the expression which is raised to the -.5 power.

^ precedes *. Since 4 x ^ -.5 means 4 * x ^ -.5, we do x^-.5 before we multiply by 4.

So it's easier to multiply both sides by 1/4, so that the left-hand side is just x^-.5, without the confusion of the 4.

Your solution illustrates why this is the best idea:

Quoting from your solution

4x^-.5 = 7 raise both sides by -1/.5

(4x^-.5)^-1/.5 = 7^-(1/.5) (this step is valid, even if it's not the best way to proceed)

4x = 0.020 (this step is in error)

The error on that last step is that (4 x ^-.5) ^ (-1 / .5) is not equal to 4x.

(4 x ^-.5) ^ (-1 / .5) =

4 ^ (-1 / .5) * (x^-.5) ^ (1 / (-.5) ) =

4^(-1/.5) * x

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Self-critique (if necessary):

I understand my error. Correct as follows below:

4 x^-.5 = 7

X^-.5 = 7*(1/4)

x^-.5 = 7/4

x^(-.5*-1/.5) which equals -.5*-2 = 7/4^(-2)

x = .3265

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Self-critique rating: ok

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Question: `qGive your solution to 14 x ^ (2/3) = 39

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Your solution:

14x^2/3 = 39

1/14*14 x^2/3 = 39 * 1/14

x^2/3 = 39/14

x^2/3*3/2 = 39/14^(3/2)

x = 4.649

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** We first multiply both sides by 1/14 to get

x^(2/3) = 39/14. The we raise both sides to the 3/2 power to get

x = (39/14)^(3/2) = 4.65. **

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `qGive your solution to 5 ( 3 x / 8) ^ (-3/2) = 9

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Your solution:

Woah…..this one is tough. I think I am going to check below, because I am not entirely sure how to do so.

confidence rating #$&*: 0

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Given Solution:

`a** multiplying both sides by 1/5 we get

(3x/8)^(-3/2) = 9/5. Raising both sides to the -2/3 power we have

3x / 8 = (9/5)^(-2/3). Multiplying both sides by 8/3 we obtain

x = 8/3 * (9/5)^(-2/3) = 1.80 **

STUDENT QUESTION

Still confused on the steps after we get down to 9/5.

INSTRUCTOR COMMENT

We take the -2/3 power of the -3/2 power:

(3x/8)^(-3/2) = 9/5. Raising both sides to the -3/2 power we have

((3x / 8)^(-3/2) ) ^ (-2/3) = (9/5)^(-2/3), so that (by the laws of exponents)

(3x / 8)^(-3/2 * (-2/3) ) = (9/5)^(-2/3). Since -3/2 * -2/3 = 1 so we have

(3x / 8)^1 = (9/5)^(-2/3), that is

3 x / 8 = (9/5)^(-2/3). Multiplying both sides by 8/3 we complete the solution.

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Self-critique (if necessary):

Ok I understand it now. These are getting tougher and tougher….(it was so easy at first haha)

5 (3x/8)^-3/2 = 9

1/5*5 (3x/8)^-3/2 = 9*1/5

3x/8 ^ -3/2 = 9/5

3x/8 ^ -3/2*-2/3 = 9/5^-2/3

3x/8 = .6758

8/3 * 3x/8 = .6758 * 8/3

X = 1.8

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Self-critique rating: ok

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Question: `qQuery problem 2. a(n+1) = a(n) + .5 n, a(0) = 2

What are a(1), a(2), a(3), a(4) and a(5)?

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Your solution:

a (n+1) = a (n) + .5

a(0) = 2

a(0+1) = a(0) + .5 (0)

a1 = a0

a1 = 2

a(1 +1) = a(1) + .5(1)

a 2 = a1 + .5

a2 = 2 + .5

a2 = 2.5

a(2+1) = a(2) + .5(2)

a3 = a2 + 1

a3 = 2.5 + 1

a3 = 3.5

a(3+1) = a(3) + .5(3)

a4 = a3 + 1.5

a4 = 3.5 + 1.5

a4 = 5

a(4+1) = a(4) + .5(4)

a5 = a4 + 4

a5 = 5+2

a5 = 7

confidence rating #$&*: 3

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Given Solution:

`a** Substituting n = 0 we get

a(0+1) = a(0) + .5 * 0 which we simplify to get

a(1) = a(0). Substituting a(0) = 2 from the given information we get

a(1) = 2.

Substituting n = 1 we get

a(1+1) = a(1) + .5 * 1 which we simplify to get

a(2) = a(1) + .5. Substituting a(1) = 2 from the previous step we get

a(2) = 2.5.

Substituting n = 2 we get

a(2+1) = a(2) + .5 * 2 which we simplify to get

a(3) = a(2) + 1. Substituting a(2) = 2.5 from the previous step we get

a(3) = 2.5 + 1 = 3.5.

Substituting n = 3 we get

a(3+1) = a(3) + .5 * 3 which we simplify to get

a(4) = a(3) + 1.5. Substituting a(3) = 3.5 from the previous step we get

a(4) = 3.5 + 1.5 = 5.

Substituting n = 4 we get

a(4+1) = a(4) + .5 * 4 which we simplify to get

a(5) = a(4) + 2. Substituting a(4) = 5 from the previous step we get

a(5) = 5 + 2 = 7. **

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `qWhat is your quadratic function and what is its value for n = 4? Does it fit the sequence exactly?

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Your solution:

I am confused as to where this question is coming from, if it’s from the last problem, I don’t understand how to get this answer???

confidence rating #$&*: 0

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Given Solution:

`a** Using points (1,2), (3,3.5) and (7,5) we substitute into the form y = a x^2 + b x + c to obtain the three equations

2 = a * 1^2 + b * 1 + c

3.5 = a * 3^2 + b * 3 + c

7 = a * 5^2 + b * 5 + c.

Solving the resulting system for a, b and c we obtain a = .25, b = -.25 and c = 2, giving us the equation

0.25•x^2 - 0.25•x + 2. **

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Self-critique (if necessary):

Now I am even more confused, I don’t know how you got those points to do the problem. I am really off on this one. I know how to do this problem if I had the coordinates to plug in. This problem is totally confusing me on how you got the coordinates, and how do you verify n = 4 from the question???????

@&

The problem in the worksheets asks you to find the values of a(1), a(2), etc., then to fit a quadratic function to the resulting points.

If you are doing the Query with your work on the assigned problems in front of you, this will be clear.

*@

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Self-critique rating:

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Question: `qQuery problem 3. f(x) = .3 x^2 - 4x + 7, evaluate at x = 0, .4, .8, 1.2, 1.6 and 2.0.

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Your solution:

.3(0)^2 - 4(.4) + 7

.048 - 1.6 + 7

5.448

.3(.8)^2 - 4(.8) + 7

.192 - 3.2 + 7

3.992

.3(1.2)^2 -4(1.2) + 7

.192 - 3.2 + 7

3.992

.3 (1.6)^2 - 4(1.6) + 7

.768 - 6.4 + 7

1.368

3(2.0)^2 - 4(20) + 7

1.2 - 8 + 7

.2

Points are (0, 7) (.4, 5.448) (.8, 3.992) (1.2, 2.632) (1.6, 1.368) (2, .2)

Y value differences are 1.552, 1.456, 1.36, 1.264, 1.168

Y value second differences are .096, .096, ……shows this is constant

X value differences are -.4, -.4, -.4…….

The graph is a linear graph.

Midpoints of x values are .2, .6, 1, 1.4, 1.8

Difference y values are 1.552, 1.456, 1.36, 1.264, 1.168

Linear function would be:

(.2, 1.552) and (1.4, 1.264)

1.264 - 1.552 = -.288

1.4-.2 = 1.2

-.288/1.2 = -.24

y = -.24x + b

1.552 = -.24(.2) + b

1.6 = b

y = -.24x + 1.6

confidence rating #$&*: 3

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Given Solution:

`a** We obtain the points

(0, 7)

(4, 5.448)

(.8, 3.992)

(1.2, 2.632)

(1.6, 1.368)

(2, .2)

y values are

7, 5.448, 3.992, 2.632, 1.368, 0.2.

Differences are 7-5.448 = -1.552, 3.992 - 2.632 = -1.456, etc. The sequence of differences is

-1.552, -1.456, -1.36, -1.264, -1.168.

The rate of change of the original sequence is proportional to this sequence of differences.

The differences of the sequence of differences (i.e., the second differences) are .096, .096, .096, .096, .096..

These differences are constant, meaning that the sequence of differences is linear..

This constant sequence is proportional to the rate of change of the sequence of differences.

The differences are associated with the midpoints of the intervals over which they occur. Therefore the difference -1.552,

which occurs between x = 0 and x = .4, is associated with x = .2; the difference -1.456 occuring between x = .4 and x =

.8 is associated with x = .6, etc..

The table of differences vs. midpoints is

}

0.2, -1.552

.6, -1.456

1, -1.36

1.4, -1.264

1.8, -1.168

This table yields a graph whose slope is easily found to be constant at .24, with y intercept -1.6. The function that

models these differences is therefore

y = 2.4 x - 1.6. **

STUDENT QUESTION:

I’m not quite sure I understand how you derived at your model y = 2.4 x - 1.6

INSTRUCTOR RESPONSE:

For example the rise between the first and second point is .096 and the run is .4, making the slope .096 / .4 = .24. You get the same slope for any pair of points (you can easily see that the x values change by the same amount each time, and if you calculate the changes in the y values you get .096 for each change, leading fairly quickly to the conclusion that the slopes are all the same).

Between 0 and 0.2 the 'rise' would therefore be 0.2 * .24 = .048. This would put the x = 0 point .48 units lower than the x = .2 point, so that at the x = 0 point the y value is -1.552 - .048 = -1.6.

The line is therefore y = 2.4 x - 1.6.

STUDENT QUESTION:

I think I may have misunderstood the directions for this one. I believe the problem statement

asked to graph the avg. rate of change vs the midpoints, which I assumed to be the slopes vs the midpoints, not the

difference between x values (which would just be the run between each point?). So my midpoint coordinates and the

resulting model do not match up to the given solution.

INSTRUCTOR RESPONSE:

Your solution was fine.

There is however a fine distinction in the wording of the question. The statement referred to a 'quadratic sequence', which isn't the same thing as a 'quadratic function'.

A quadratic sequence is a sequence of numbers which can be obtained by evaluating a quadratic function. Once the numbers are obtained, the function is effectively discarded, and the sequence stands on its own.

The function yields the sequence in the manner you outline:

f(x) = 0.6x - 4

f(0.2) = 0.6(0.2) - 4 = -3.88

f(0.6) = 0.6(0.6) - 4 = -3.64

f(1.0) = 0.6(1.0) - 4 = -3.4

f(1.4) = 0.6(1.4) - 4 = -3.16

f(1.8) = 0.6(1.8) - 4 = -2.92

The original function f(x) is at this point forgotten. The sequence is -3.88, -3.64, -3.4, -3.16, -2.92, ...

A sequence is numbered by integers, so that these values could now be referred to as a(1), a(2), a(3), a(4) and a(5).

Since a sequence is numbered by consecutive integers, the rates at which a sequence changes are just the differences in consecutive members of the sequence.

You might question why we want to forget the origins of the sequence. The reason is that the sequence is just simpler, and has certain properties and behaviors that are better understood if we treat it as such.

Having done so we can relate our results for the sequence back to the original function and its rate-of-change behavior.

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Self-critique (if necessary):

I noticed that the function I have is different than the one given??? What have I mixed up???

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Self-critique rating:

@&

Your solution appears to be correct.

*@

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Question: `qQuery problem 4. f(x) = a x^2 + b x + c

What symbolic expression stands for the average slope between x = h and x = k?

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Your solution:

F(h) = ah^2 + bh + c

F(k) = ak^2 + bh + c

[(ak^2 + bk + c) - (ah^2 + bh + c)] / k-h

a k^2 - ah^2 + bk - bh / k - h

a (k2-h^2) + b(k-h) / k - h

a (k-h) + b / k - h

a + b

confidence rating #$&*: 2

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Given Solution:

`a** The average slope is rise / run = [ f(k) - f(h) ] / (k - h) = ( a k^2 + b k + c - ( a h^2 + b h + c) ) / ( k - h).

We simplify this to get

ave slope = ( a ( k^2 - h^2) + b ( k - h) ) / ( k - h), which we write as

ave slope = ( a ( k-h) ( k+h) ) + b ( k - h) ) / (k - h).

k - h is a factor of the numerator so we have the final form

ave slope = a ( k + h) + b. **

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Self-critique (if necessary):

I understand what I missed. In order to simplify a (k^2 - h^2) you need to have a (k-h)(k+h). Notes taken and updated.

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Question: `qQuery last asst before test 1, problem 1.Give your solution to x ^ 3 / 17 = 58

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Your solution:

x^3 / 17 = 58

x^3 = 986

x = 9.95

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** The solution to x^3 / 17 = 58 is found by first multiplying both sides by 17 to get

x^3 = 58 * 17

then taking the 1/3 power of both sides, obtaining

(x^3)^(1/3) = (58 * 17)^(1/3) or

x = 9.95, approx..

COMMON ERROR:

If you interpret the equation as x^(3/17) = 58 you will get solution x = 58^(17/3) = 9834643694. However this is not

the solution to the given equation

To interpret x ^ 3 / 17 you have to follow the order of operations. This means that x is first cubed (exponentiation

precedes multiplication or division) then divided by 17. If you introduce the grouping x^(3/17) you are changing the

meaning of the expression, causing 3 to be divided by 17 before exponentiation. **

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Self-critique (if necessary):

ok

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Question: `qGive your solution to (3 x) ^ -2 = 19

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Your solution:

3x^-2 = 19

3x = .229

X = .0763

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** (3x)^-2 = 19 is solved by taking the -1/2 power of both sides, or the negative of the result:

((3x)^-2)^(-1/2)) = 19^(-1/2) gives us

3x = 19^(-1/2) so that

x = [ 19^(-1/2) ] / 3 = .0765 or -.0765. **

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Self-critique (if necessary):

ok

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Question: `qGive your solution to 4 x ^ -.5 = 7

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Your solution:

4x^-.5 = 7

x^-.5 = 7/4

x ^(-.5 * -1/.5)= -.5*-.2 = 7/4 ^ -2

x = 3.265

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** to solve the equation we first multiply both sides by 1/4 to get

x ^ -.5 = 7 / 4. Then we raise both sides to the -2 power:

(x^-.5)^-2 = (7/4)^-2 so

x = .327 approx **

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Self-critique (if necessary):

ok

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Question: `qGive your solution to 14 x ^ (2/3) = 39

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Your solution:

14 x^2/3 = 39

x^2/3 = 39/14

x = 39/14^3/2

x = 4.649

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** We first multiply both sides by 1/14 to get

x^(2/3) = 39/14. The we raise both sides to the 3/2 power to get

x = (39/14)^(3/2) = 4.65. **

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Self-critique (if necessary):

ok

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: `qGive your solution to 5 ( 3 x / 8) ^ (-3/2) = 9

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Your solution:

5 (3x / 8) ^ -3/2 = 9

3x/8 ^ -3/2 = 9/5

3x/8 = 9/5 ^ -2/3

3x / 8 = .6758

X = .6758 * 8/3

x = 1.8

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** multiplying both sides by 1/5 we get

(3x/8)^(-3/2) = 9/5. Raising both sides to the -2/3 power we have

3x / 8 = (9/5)^(-2/3). Multiplying both sides by 8/3 we obtain

x = 8/3 * (9/5)^(-2/3) = 1.80 **

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Self-critique (if necessary):

ok

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Self-critique rating:

ok

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Self-critique (if necessary):

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Self-critique rating:

ok

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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#