assign 23

This document is being posted approximately one week late. It was posted when it was originally submitted, but due to an error it went to an incorrect location.

course mth 158

I see the submit work form is working again.

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.

023. `* 23

*********************************************

Question: * 3.3.16 (was 3.2.6). Key pts and behavior: far left incr, far right incr, zeros at -10, 5, 0, 5, peaks at (-8,-4), (-2, 6), (2, 10). Local min, max among listed points.List the intervals on which the function is decreasing.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Q 3.3.16 reads: List the interval(s) on which f is decreasing.

The line is decreasing on the open intervals (-10, -8) or for -10 < x < -8 and on the open intervals (-2, 0) or for -2 < x < 0 and again on the open intervals (2, 5) or 2 < x < 5

confidence rating:

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

.............................................

Given Solution:

* * The function decreases until reaching the local min at (-8, -4), then increases until reaching the local max at (-2, 6).

The function then decreases to its local min at (5, 0), after which it continues increasing.

So the graph is decreasing on (-infinity, -8), on (-2, 0) and on (2, 5). **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I didn’t state the local maxima and local minima, but for that particular problem, they are:

local minima (-8, -4) (0, 0) (5, 0)

local maxima (-2, 6) (2, 10)

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: * 3.3.22 (was 3.2.12). Piecewise linear (-3,3) to (-1,0) to (0,2) to (1,0) to (3,3).Give the intercepts of the function. Give the domain and range of the function. Give the intervals on which the function is increasing, decreasing, and constant. Tell whether the function is even, odd or neither.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph for problem 3.3.22

a) The intercepts, if any

x-axis (-1, 0) y-axis (0, 2) and again on the x-axis (1, 0)

b) The domain and range are:

D: [-3, 3] or all possible values of x {x| -3 <= x <= 3}

R: [0, 3] or all possible values of y {y| 0 <= y <= 3}

c) The intervals on which it is increasing, decreasing, or constant

increasing from (-1, 0) to (0, 2) and from (1, 0) to (3, 3) on the open intervals (-1, 0) and (1, 3)

decreasing from (-3, 3) to (-1, 0) and from (0, 2) to (1, 0) on the open intervals (-3, -1) and (0, 1)

there are no constants

d) Whether it is even, odd, or neither

By looking at the graph, I can see that it is symmetric with respect to the y-axis, which makes it an even function.

confidence rating:

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

.............................................

Given Solution:

* * The function intersects the x axis at (-1, 0) and (1, 0), and the y axis at (0, 2).

The function decreases from (-3,3) to (-1,0) so it is decreasing on the interval (-3, -1).

The function decreases from (0, 2) to (1, 0) so it is decreasing on the interval (0, 1).

The function increases from (-1,0) to (0, 2) so it is increasing on the interval (-1, 0).

The function ioncreases from (1,0) to (3, 3) so it is increasing on the interval (1, 3).

The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3].

The range of the function is the set of possible y values, so the range is 0 <= y <= 3, written as the interval [0, 3].

The function is symmetric about the y axis, with f(-x) = f(x) for every x (e.g., f(-3) = f(3) = 3; f(-1) = f(1) = 0; etc.). So the function is even. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I didn’t show the function f(-x) = f(x) to show the graph was even. I’m not sure if it would be enough on a test to just say from the look of the graph, it is symmetric and I should probably remember to put the function as proof of the symmetry.

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: * 3.3.28 (was 3.2.18). Piecewise linear (-3,-2) to (-2, 1) to (0, 1) to (2, 2) to (3, 0)Give the intercepts of the function. Give the domain and range of the function. Give the intervals on which the function is increasing, decreasing, and constant. Tell whether the function is even, odd or neither.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph for 3.3.28

a) The intercepts

x-axis (-2.3, 0) y-axis (0, 1) and again on the x-axis (3, 0)

b) The domain and Range

D: [-3, 3] all possible values of x {x| -3 <= x <= 3}

R: [-2, 2] all possible values of y {y| -2 <= y <= 2}

c) The intervals on which it is increasing, decreasing, or constant

increasing from (-3, -2) to (-2, 1) and from (0, 1) to (2, 2) on the open intervals (-3, -2) and (0, 2)

decreasing from (2, 2) to (3, 0) on the open intervals (2, 3)

constant from (-2, 1) to (0, 1) on the open intervals (-2, 0)

d) Whether it is even, odd, or neither

I’m not really sure how to replace x with –x in the case of a graph when there is no equation.

There are some –x to some x, but not all, so I would say that the graph is not an even function. To be an odd function, it would have to be symmetric with respect to the origin, which clearly it is not, so I would say it is neither even nor odd.

confidence rating:

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

.............................................

Given Solution:

* * The function intersects the x axis at (-2.25, 0) and (3, 0), and the y axis at (0, 1).

The function decreases from (2,2) to (3,0) so it is decreasing on the interval (2,3).

The function increases from (-3,-2) to (-2, 1) so it is increasing on the interval (-3, -2).

The function ioncreases from (0, 1) to (2, 2) so it is increasing on the interval (0, 2).

The function value does not change between (-2, 1) and (0, 1) so the function is constant on (-2, 0).

The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3].

The range of the function is the set of possible y values, so the range is -2 <= y <= 2, written as the interval [-2, 2].

The function is not symmetric about the y axis, with f(-x) not equal to f(x) for many values of x; e.g., f(-3) is -2 and f(3) is 0. So the function is not even.

x = 3 shows us that the function is not odd either, since for an odd function f(-x) = -f(x) and for x = 3 this is not the case. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I think the give solution meant that it intersected at the x axis at (-2.3, 0) instead of (-2.25, 0)

There are no discrepancies.

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: * 3.3.32 (was 3.2.24). sine-type fn (-pi,-1) to (0, 2) to (pi, -1).At what numbers does the function have a local max and what are these local maxima?At what numbers does the function have a local min and what are these local minima?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Problem 3.3.32 states

a) The numbers, if any, at which f has a local maximum. What are these local maxima?

There is one local maxima, at the point (0, 1)

b) The numbers, if any, at which f has a local minimum. What are these local minima?

There are two, one at the point (-pi, -1) and the other (pi, -1)

confidence rating:

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

.............................................

Given Solution:

* * Local maximum is (0,1)

Local minimum are (-pi,-1) and (pi,-1) **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

There were no discrepancies.

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: * 3.3.76 / 46 (was 3.2.36) (f(x) - f(1) ) / (x - 1) for f(x) = x - 2 x^2What is your expression for (f(x) - f(1) ) / (x - 1) and how did you get this expression? How did you use your result to get the avg rate of change from x = 1 to x = 2, and what is your value? What is the equation of the secant line from the x = 1 point to the x = 2 point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(f(x) - f(1) ) / (x - 1) for f(x) = x - 2 x^2

average rate of change from x = 1 to x = 2

(f(x) - f(1) ) / (x - 1)

= -6 -1 / 2 -1 = - 15/2

I really don’t know what I’m doing on this problem. Every time I think I have a grasp on the idea, it slips away. I need to practice doing these a little more.

confidence rating:

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

.............................................

Given Solution:

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

The above line had a misleading typo. It has been corrected.

This factors into (2x + 1) ( -x + 1).

Since -x + 1 = - ( x - 1) we obtain

(f(x) - f(1) ) / ( x - 1) = (2x + 1) ( -x + 1) / (x - 1) = (2x + 1) * -1 = - (2x + 1).

A secant line runs from one point of the graph to another. The secant line here runs from the graph point (1, f(1) ) to the graph ponit ( x , f(x) ), and the expression we have just obtained is the slope of the secant line.

For x = 2 the expression -(2x + 1) gives us ( f(2) - f(1) ) / ( 2 - 1), which is the slope of the line from (1, f(1) ) to (2, f(2)) .

-(2 * 2 + 1) = -5, which is the desired slope.

The secant line contains the point (1, -1) and has slope 5. So the equation of the secant line is

(y - (-1) ) = -5 * (x - 1), which we solve to obtain

y = -5 x + 4. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I can follow along somewhat. I need more practice.

------------------------------------------------

Self-critique Rating: 0

*********************************************

Question: * 3.3.36 / 50 (was 3.2.40). h(x) = 3 x^3 + 5. Is the function even, odd or neither? How did you determine algebraically that this is the case?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

h(x) = 3 x^3 + 5 replace x with –x to see if it is an even function

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

3(-1)^3 + 5 which is not the same as 3x^3 + 5 so it is not an even function

confidence rating:

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

.............................................

Given Solution:

* * h(x) = 3x^3 +5

h(-x) = 3-x^3 +5

= -3x^3 + 5

h(x) is not equal to h(-x), which means that the function is not even.

h(x) is not equal to -h(-x), since 3 x^3 + 5 is not equal to - ( -3x^3 + 5) = 3 x^3 - 5. So the function is not odd either. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I am still a little fuzzy about solving these. I need more practice. I think I have the same answer as the given solution.

------------------------------------------------

Self-critique Rating: 3

Good work overall, but be sure to see my note(s) and let me know if you have specific questions.

"