Assignment 24 34

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course Mth 158

11 7/11

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.

024. `* 24

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Question: * 3.4.14 (was 3.3.6). Concave down then concave up.

Does this graph represent a constant, linear, square, cube, square root, reciprocal, abs value or greatest integer function?

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Your solution:

The graph represents a cube function.

confidence rating #$&*: 3

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Given Solution:

* * A linear function, represented most simply by y = x, has no curvature.

A quadratic function, represented most simply by y = x^2, has a parabolic graph, which is either concave up or concave down.

A cubic function, represented most simply by y = x^3, has a graph which changes concavity at the origin. This is the first power which can change concavity.

A square root function, represented most simply by y = sqrt(x), have a graph which consists of half of a parabola opening to the right or left. Its concavity does not change.

A reciprocal function, represented most simply by y = 1 / x, has a graph with a vertical asymptote at the origin and horizontal asymptotes at the x axis, and opposite concavity on either side of the asymptote.

An absolute value function, represented most simply by y = | x |, has a graph which forms a 'V' shape.

The greatest integer function [[ x ]] is a 'step' function whose graph is made up of horizontal 'steps'.

The concavity of the reciprocal function changes, as does the concavity of the cubic function. The cubic function is the only one that matches the graph, which lacks the vertical and horizontal asymptotes. **

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question: * 3.4.20 (was 3.3.12).

Does your sketch of f(x) = sqrt(x) increase or decrease, and does it do so at an increasing or decreasing rate?

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Your solution:

The graph increases at an increasing rate.

confidence rating #$&*:

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Given Solution:

* * y = sqrt(x) takes y values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4.

sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2.

The change in y from the x = 0 to the x = 2 point is about 1.414; the change from the x = 2 to the x = 4 point is about .586. The y values therefore increase, but by less with each 'jump' in x.

So the graph contains points (0,0), (2, 1.414) and (4, 2) and is increasing at a decreasing rate. **

What three points did you label on your graph?

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Self-critique (if necessary): I don’t understand where the plug in numbers came from? Did you just graph the square root function and choose random numbers that would be on there? How do you know what numbers would fall on that line? I understand that a square root function has a intercept of (0, 0) and it increases at the interval of (0, infinity). I am confused on how you take that information and apply it to a problem without numbers or equations.

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Self-critique Rating: 1

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This is very similar to the problem you did in the graphing document of the Initial Activities.

In any case, you could pick your own x values. If you pick a handfull of x values and evaluate the function, you will come to the same conclusion.

I picked numbers whose square roots you should know.

Another good way to pick numbers for this function is to pick perfect squares. You could let x = 0, 1, 4, 9, 16, obtaining y = 0, 1, 2, 3, 4. It is clear that this function is increasing at a decreasing rate.

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Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0.

Given the intercepts, domain and range of the function

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Your solution:

The graph y=2x+5 has an intercept of 5 and a slope of 2. The x intercept happens when y is equal to 0 so you must solve the equation.

2x+5=0

subtract 5

2x= -5

Divide by 2

= - 5/2

So -5/2 is our x and the y=0

If x=-3 then you plug -3 into the equation for x.

2(-3)+5= -1

So x= -3 and y= -1

Our three points for the function would be (-3, -1) (-5/2, 0) and (0, 5)

I have no idea what to do after this part. I graphed the function on graph paper but do not know how to figure out the domain and range.

confidence rating #$&*: 1

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Given Solution:

* * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at

2x + 5 = 0 or x = -5/2.

Since this x value is in the interval from -3 to 0 it is part of the graph.

At x = -3 we have y = 2 * (-3) + 5 = -1.

So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5).

The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5).

For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin.

For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). **

STUDENT COMMENT

I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet.

INSTRUCTOR RESPONSE

The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity).

The range is the set of all possible values of f(x), where x can be any number in the domain.

On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5.

For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value.

Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range.

The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3):

The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3):

The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively.

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Self-critique (if necessary): I don’t understand in your solution why the line increases up to (0, 5) but does not include it? Also, I do not know where the other line came from (the blue line)? I had the pink/red line graphed.

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This function has three different definitions, each applying for different values of x.

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The function doesn't get to the value 5, because at x = 0 the function no longer has the definition 2 x + 5. It does have the definition right up to x = 0, which is why the value goes right up to y = 5. But when x is zero, the function's value suddenly 'jumps' down to -3.

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Starting just past x = 0 your graph would have to agree with the graph of y = -5 x.

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I got lost at this point in your explanation:

The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5).

For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin.

For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). **

I don’t think I quite understand the -3 < x < 0?

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The interval (-3, 0) can be written as the inequality -3 < x < 0.

For these x values, and for no others, the function is defined by the formula y = 2x + 5.

So the function follows a line from (-3, -1) to (0, 5). However this definition does not apply at x = -3 or at x = 0, since it applies only for the open interval -3 < x < 0. Thus the entire line segment is included, except for its endpoints at (3, -1) and (0, 5).

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Self-critique Rating: 1

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Self-critique (if necessary):

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