assign 32

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.

032. * 32

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Question: * 5.1.20 / 7th edition 4.2.20 (was 4.2.10). If f(x)= (x^2-5) / x^3 a polynomial?

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Your solution:

I think the question is asking if it is a polynomial function.

It is the ratio of two polynomials, so it is not a polynomial function.

confidence rating: 3

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Given Solution:

* * This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3.

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating: 3

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Question: * 5.1.20 / 7th edition 4.2.40 (was 4.2.30). If a polynomial has zeros at x = -4, 0, 2 the what is its minimum possible degree, and what form will the polynomial have?

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Your solution:

Zeros are x = -4, 0, 2

This would mean that 0 and 2 are factors of x = -4

To set this up you would choose a leading coefficient. I will use a = 1

(x – (-4)) = a(x+4)(x-0)(x-2)

= x^2+x^2-2x+4x-0+4x-8

=2x^2 + 6x -8 (which is a quadratic equation)

the degree is 2

confidence rating: 3

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Given Solution:

* * The factors of the polynomial must include (x - (-4) ) = x + 4, x - 0 = x, and x - 2. So the polynomial must be a multiply of (x+4)(x)(x-2).

The general form of the polynomial is therefore

f(x)=a(x+4)(x-0)(x-2).

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Self-critique (if necessary):

I don’t know if the given solution is the same as my solution. Did I need to work it like that or could I have just said f(x)=a(x+4)(x-0)(x-2)?

That would have been OK, but you should be prepared to multiply the polynomial out, as you did. Some questions will ask you to do that.

Do note that you multiplied incorrectly. The product of (x + 4) ( x - 2) ( x - 0) is

(x^2 + 2 x - 8) * (x - 0), where we've first multiplies (x + 4) ( x - 2). Then since x - 0 is just x, we get

x^3 + 2 x^2 - 8 x.

It's best to multiply one step at a time, using the distributive law carefully.

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Self-critique Rating: 3

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Question: * 5.1.42 / 7th edition 4.2.52 (was 4.2.40). What are the zeros of the function f(x)=(x+sqrt(3))^2 (x-2)^4 and what is the multiplicity of each?

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Your solution:

f(x)=(x+sqrt(3))^2 (x-2)^4

The zeros are x + sqrt(3) = 0 or x-2 = 0

so x = -sqrt(3) and x = 2

The multiplicity of x = -sqrt(3) is 2

The multiplicity of x = 2 is 4

confidence rating: 3

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Given Solution:

* * f(x) will be zero if x + sqrt(3) = 0 or if x - 2 = 0.

The solutions to these equations are x = - sqrt(3) and x = 2.

The zero at x = -sqrt(3) comes from (x + sqrt(3))^2 so has degree 2.

The zero at x = 2 comes from (x-2)^4 so has degree 4.

For each zero does the graph touch or cross the x axis?

In each case the zero is of even degree, so it just touches the x axis. Near x = -sqrt(3) the graph is nearly a constant multiple of (x+sqrt(3))^2. Near x = 2 the graph is nearly a constant multiple of (x - 2)^4.

What power function does the graph of f resemble for large values of | x | ?

If you multiply out all the terms you will be a polynomial with x^6 as the highest-power term, i.e., the 'leading term'.

For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function x^6. **

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Self-critique (if necessary):

I didn’t see the question of the power function until now. I am still a little unsure how to solve this. I have been reading ch5 again and the way I understand the power function at large values of |x| is this:

There are only 4 types.

1. It opens up like that of a parabola when the power function is an even number (2, 4, 6….) and the leading number is greater than 0.

2. It open down when the power function is even and the leading number is less than 0.

3. It looks like an asymmetric graph with respect to the origin with the points (1,1) and (-1,-1) when the power function is odd and the leading number is greater than 0.

4. It looks like number 3 only it has been flipped with points (-1, -1) and (1, -1) when the power function is odd and the leading number is less than 0.

If my interpretation is correct, then how do you multiply out the terms to get the highest power term? It seems like I have more trouble with the basics like (multiplying out the terms, and long division of polynomials) than grasping this.

This is good. See the original Query document for a response to your question, along with some explanatory graphs.

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Self-critique Rating: 3

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Question: * 5.1.61 / 7th edition 4.2.62 (was 4.2.50). f(x)= 5x(x-1)^3. Give the zeros, the multiplicity of each, the behavior of the function near each zero and the large-|x| behavior of the function.

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Your solution:

f(x)= 5x(x-1)^3

Zeros are 5x = 0 or x-1 = 0

So x = 0 and x = 1

The multiplicity of x = 0 is 1

The multiplicity of x = 1 is 3

Since it is an odd degree, the graph crosses the x axis.

Behavior near each zero

Near x = 0 5x(0-1)^3 = 0

Near x=1 5(0) (x-1)^3 = 0

The large -|x| behavior of the function still is not clear to me here.

confidence rating: 2

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Given Solution:

* * The zeros occur when x = 0 and when x - 1 = 0, so the zeros are at x = 0 (multiplicity 1) and x = 1 (multiplicity 3).

Each zero is of odd degree so the graph crosses the x axis at each.

If you multiply out all the terms you will be a polynomial with 5 x^4 as the highest-power term, i.e., the 'leading term'.

For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function 5 x^4.

What is the maximum number of turning points on the graph of f?

This is a polynomial of degree 4. A polynomial of degree n can have as many as n - 1 turning points. So this polynomial could possibly have as many as 4 - 1 = 3 turning points.

Give the intervals on which the graph of f is above and below the x-axis

this polynomial has zeros at x = 0 and x = 1.

So on each of the intervals (-infinity, 0), (0, 1) and (1, infinity) the polynomial will lie either wholly above or wholly below the x axis.

If x is a very large negative or positive number this fourth-degree polynomial will be positive, so on (-infinity, 0) and (1, infinity) the graph lies above the x axis.

On (0, 1) we can test any point in this interval. Testing x = .5 we find that 5x ( x-1)^3 = 5 * .5 ( .5 - 1)^3 = -.00625, which is negative.

So the graph lies below the x axis on the interval (0, 1).

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Self-critique (if necessary):

I just saw where it asks the question…what are the maximum numbers of turning points of the graph? I don’t see how the degree is 4 in this particular problem. I know that if it is 4, then the turning points would be n-1 making it 3 turning points.

f(x)= 5x(x-1)^3

 

Expanding we get

5x(x-1)^3 = 5 x ( x - 1) ( x - 1) ( x - 1) = 5 x ( x^2 - 2 x + 1) ( x - 1), where we have multiplied (x - 1) ( x - 1).

Now

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

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

= x^3 - 2 x^2 + x - x^2 + 2 x - 1

= x^3 - 3 x^2 + 3 x - 1,

so our product becomes

 

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

= 5 x ( x^3 - 3 x^2 + 3 x - 1 )

= 5 x^4 - 15 x^3 + 15 x - 5.

For this particular problem, you don't actually have to do all the multiplication but on others you might. The point here is to understand why the degree of the polynomial is 4.

The graph doesn't have to have 3 turning points. For example y = x^4 is a polynomial of degree 4, but has only one turning point.

The graph of a degree-4 polynomial has at most 3 turning points, but it can have fewer.

Similarly the graph can have as many as 4 zeros, but it can have fewer. For example y = x^4 + 1 has no zeros, while y = x^4 - 1 has two zeroes and the graph of y = (x - 2) ( x - 1) ( x + 1) ( x + 3) has four zeros.

Then there is the question of where the graph lies above or below the x axis.

We know the graph cross at x = 0 and x = 1 so we divide the graph into 3 sections.

(-inf to 0) (0 to 1) (1 to inf)

Then we set up a table with the information:

Number chosen -1 ½ 2

Value of f f(-1) = 40 f(1/2) = - 5/16 f(2) = 10

Location of the graph above x axis, below x axis, above x axis

Point on graph (-1, 40) (1/2, -5/16) (2, 10)

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Self-critique Rating: 3

Once more you've asked excellent questions.

See my notes (and the newly edited Query document) and let me know if you have additional questions.