QUERY 10

course MTH 271

10/12, 1440

z~nķ黳rfұፚ⏷sassignment #010

010. `query 10

Applied Calculus I

10-10-2009

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13:40:46

What is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.

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RESPONSE -->

A polynomial with zeros at -3, 4, and 9 crosses the x-axis at (-3,0), (4,0), and (9,0). To determine the polynomial equation, we find the equation for each zero.

x=-3....x+3=0 or (x+3)

x=4....x-4=0 or (x-4)

x=9....x-9=0 or (x-9)

We then multiply these equations to find the polynomial equation.

(x+3)(x-4)(x-9)

(x^2 -x -12)(x-9)

x^3 -x^2 -12x -9x^2 + 9x + 108

x^3-10x^2 -3x + 108

The graph looks like two parabolae have been combined with the first opening downward and the second opening upward.

confidence rating: 2

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13:43:07

A polynomial with zeros at -3, 4 and 9 must have factors (x + 3), ( x - 4) and (x - 9), and so must contain (x+3) (x-4) (x-9) as factors.

These factors can be multiplied by any constant. For example

8 (x+3) (x-4) (x-9),

-2(x+3) (x-4) (x-9) and

(x+3) (x-4) (x-9) / 1872

are all polynomizls with zeros at -3, 4 and 9.

If -3, 4 and 9 are the only zeros then (x+3), (x-4) and (x-9) are the only possible linear factors.

It is possible that the polynomial also has irreducible quadratic factors. For example x^2 + 3x + 10 has no zeros and is hence irreducible, so

(x+3) (x-4) (x-9) (x^3 + 3x + 10) would also be a polynomial with its only zeros at -3, 4 and 9.

The polynomial could have any number of irreducible quadratic factors. **

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RESPONSE -->

Instead of providing for the many possibilities of polynomials that could be produced with the given zeros, I solved for the polynomial equation and the corresponding graph.

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Self-critique Rating:ent: 2

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13:45:41

1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin fn)

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RESPONSE -->

A. right-hand limit: -2

B. left-hand limit: -2

C. limit: -2

confidence rating: 2

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13:47:14

What are the three limits for your function (if a limit doesn't exist say so and tell why)?

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RESPONSE -->

The three limits for the sine function all equal -2.

confidence rating: 2

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13:47:55

Imagine walking along the graph from the right. As you approach the limiting x value, your y 'altitude' gets steadier and steadier, approaching closer and closer what value? You don't care what the function actually does at the limiting value of x, just how it behaves as you approach that limiting value.

The same thing happens if you walk along the graph from the left. What does you y value approach?

Is this the value that you approach as you 'walk in' from the right, as well as from the left? If so then it's your limit. **

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RESPONSE -->

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Self-critique Rating:ent: 3

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13:51:08

1.5.22 (was 1.5.20 right-, left-hand limits and limit (discont at pt)

What are the three limits for your function (if a limit doesn't exist say so and tell why)?

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RESPONSE -->

A. right-hand limit=0

B. left-hand limit=2

C. limit= Limit does not exist b/c the one-sided limits are not equal.

confidence rating: 2

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13:51:20

STUDENT RESPONSE:

The three limits are a. 0 b. 2 c. no limit The limit doesn't exist because while the function approaches left and right-handed limits, those limite are different.

INSTRUCTOR COMMENT:

That is correct.

ADVICE TO ALL STUDENTS:

Remember that the limit of a function at a point depends only on what happens near that point. What happens at the point iself is irrelevant to the limiting behavior of the function as we approach that point.

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RESPONSE -->

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Self-critique Rating:ent: 3

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13:53:01

1.5.30 (was 1.5.26 lim of (x+4)^(1/3) as x -> 4

What is the desired limit and why?

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RESPONSE -->

Using direct substitution, we are able to find the limit as x approaches 4.

(x+4)^(1/3)

(4+4)^(1/3)

(8)^(1/3)

2

confidence rating: 3

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13:53:08

The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2. **

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RESPONSE -->

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Self-critique Rating:ent: 3

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14:02:22

1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x -> 1

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RESPONSE -->

First, we must factor (x^3 - 1). Since the limit is approaching -1, we suspect that (x + 1) will be a factor.

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

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

(x^2 - x + 1) (Like factors have been divided out)

Now, we use direct substitution:

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

(1 + 1 + 1)

3

The limit as x approaches -1 is 3.

confidence rating: 2

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14:04:16

As x -> 0 both numerator and denominator approach 0, so you can't tell just by plugging in numbers what the limit will be.

If you factor x^3-1 into (x-1)(x^2+x+1) you can reduce the fraction to (x-1)(x^2+x+1) / (x-1) = x^2 + x + 1, which is equal to the original function for all x except 1 (we can't reduce for x = 1 because x-1 would be zero, and we can't divide by zero).

As x -> 1, x^2 + x + 1 -> 3. It doesn't matter at all what the function does at x = 1, because the limiting value of x never occurs when you take the limit--only x values approaching the limit count. 3 is therefore the correct limit. **

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RESPONSE -->

It seems the problem given by the query for 1.5.48 and that given by the book differ slightly, yet still have the same limit.

As given by the book, the problem is:

(x^3 -1) / (x+1), limit as x approaches -1

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Self-critique Rating:ent: 3

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14:06:09

1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6%

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RESPONSE -->

r=.06

A=2000 ( 1 + (.06/4))^40

A=3628.04

The limit of A as r approaches 6% is 3628.04.

confidence rating: 2

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14:07:26

$1000 *( 1+.06 / 40)^40 = 1061.788812.

Since this function changes smoothly as you move through r = .06--i.e., since the function is continuous at r = .06--this value will be the limit. **

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RESPONSE -->

Again, there is a difference between the problem given by the query and by the book. The book problem is:

1.5.70

A=2000 ( 1 + (r/4))^40

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Self-critique Rating:ent: 2

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14:07:42

Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

None noted.

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Self-critique Rating:ent: 3

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&#This looks very good. Let me know if you have any questions. &#