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

 

 

Question: *  5.1.24 / 5.1.20 / 7th edition 4.2.20 (was 4.2.10). Is f(x)= (x^2-5) / x^3 a polynomial?

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

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

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *  5.1.44 / 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?

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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 multiple of (x+4)(x)(x-2).

 

The general form of the polynomial is therefore

 

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

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *  Extra Problem / 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?

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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. **

 

STUDENT QUESTION:

 

 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.

 

INSTRUCTOR RESPONSE

 

Your list of the behaviors of the power functions is correct, as well expressed.  Your list applies only to positive integer powers,  That is appropriate here because we are dealing with polynomial functions, whose zeroes involve only positive integer powers.

 

The function given in this problem involves even powers of both factors.  To illustrate both even and odd powers, we're going to analyze here the function h(x) = (x+sqrt(3))^2 (x-2)^3 rather than the given function f(x)=(x+sqrt(3))^2 (x-2)^4:

h(x)=(x+sqrt(3))^2 (x-2)^3 has zeroes when x + sqrt(3) = 0 and when x - 2 = 0. 

The graph of h(x) is depicted in the figure below:

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *  Extra Problem / 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.

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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).

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: