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course Mth 174
6/26 7Professor Smith,
Would it be possible to switch this class to an Audit class so I can just receive a pass or fail at the end of the semester, and omit testing?
I need to finish up the two physics courses for transfer into tech, and would like to keep access to the material in MTH 176 and just hand in the assignments for review.
I wanted to ask if this would be possible rather than withdrawing and ending up with a W before the W date.
Let me know if this would be possible.
I will copy this into the official question forum.
Along are my answers to section 6.1 questions." "Calculus 2
Section 6.1
1) *6.1.2 Sketch and describe two functions F such that F' = f when f is the line going through (1,0) and (0,1).
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f'(x), from the interval of x from (0,1) decreases at a constant rate from the x,y point (1,0) to the x,y point (0,1).
The slope between these points is ('dy / 'dx) = ( 0 - 1) / (1 - 0) = -1.
The slope of F = F' is -1.
The antiderivative of f' is then equal to : x + C | from (1 to 0)
C can be any y point of value, and the resulting line will be a horizontal line starting at x + C = -1 + C
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2) *6.1.4 Sketch and describe two functions F such that F' = f when f(x) = |x|
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The antiderivative = (1/2) * x^2 + C
The graph of F will be a parabola that opens up with the vertex at point C.
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3) 6.1.6 Estimate f(x) for x = 2,4,6 when f(0) = 50 and f '(0)= 17, f '(2)= 15, f '(4) =10, and f '(6) = 2.
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The graph of f'(x) starts at the point (0,17) and then decreases at a increasing rate to the point (6,2)
I think the area under the f'(x) graph represents the change or addition of F over the respected interval.
For the interval of x from (0,2) f'ave = (17 + 15)/2 = 16, and the area is fAve * 'dx = 32
so the point will will be (2,(50 + 32)) = (2,82).
Next point of F(4) = f'Ave* 'dx = ((15 + 10) / 2) * 2 = 25
So the point will be ( 4, 82 + 25) = (4, 107)
The next point at F(6) = (10 + 2) / 2 * 'dx = 12
so the point at F(6) = (6, 107 +12) = (6,119)
The graph of F increases at a decreasing rate with points as follows
0,50
2,82
4,107
6,119
x coordinate, y coordinate.
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4) *6.1.8 Let dP/dt be described by the piecewise defined function which follows. When t <= 2, dP/dt = -1. When 2 <= t <= 4, dP/dt = t - 3. When t >= 4, dP/dt = 1. Sketch dP/dt and using P(0) = 2 use dP/dt to find P(t) when t = 1,2,3,4,5.
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1, -1
2, -1
3, 0
4, 1
5, 1
tVavlue, P Value.
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5) f(x) is a function defined on the interval 0 <= x <= 7 with the following characteristics:
f '(x) = 1 for all x on the interval (0,2),
f '(x) = -1 for all x on the interval (2,3),
f '(x) = 2 for all x on the interval (3,4),
f '(x) = -2 for all x on the interval (4,6),
f '(x) = 1 for all x on the interval (6,7)
f(3) = 0
What is the integral of f ' (x) over the interval 0 <= x <= 7?
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intergral (f'x,dx,0,7) = sum of areas in interval
Sum of areas = (1*2) + (-1*1) + (2*1) + (-2*2) + (1*1) = 0
the intergral is = 0.
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Describe your graph of f(x), indicating where it is increasing and decreasing and where it is concave up, where it is straight and where it is concave down.
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The graph of f(x) increases at a constant rate from x = (0,2)
The graph of f(x) decreases at a constant rate from x (2,3) but takes the value of f(3) = 0
The graph of f(x) increases at a constant rate from x (3,4)
The graph of f(x) decreases at a constant rate from x (4,6)
The graph of f(x) increases at a constant rate from x (6,7)
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What are the following values:
f(2) = 2
f(3) = 0, hole at 1
f(4) = 3
f(6) = -1
f(7) = 0
Awnsers are to the left of ""=""
Is the graph of f(x) continuous?
no the graph has a hole at x = 3 of f(3).
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6) *6.1.10 Let f (x)= x^3 - 3x^2 and F'(x) = f(x). Sketch or graph f(x) (with a graphing calculator if necessary, though you should be able to construct the graph easily) and referring only to the graph of f(x), give the critical points, explain which are local maxima/minima, and which are neither. Also sketch a possible graph of F(x).
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f'(x) = 3x^2 - 6x
0 = 3x^2 - 6x, x = 2.
slope is zero at f(2) = -4
Minimum at the point (2,-4)
f''(x) = 0 will give the inflection point
f''(x) = 6x - 6
x = 1.
f(1) = -2
inflection at (1,-2)
The graph of f(x) goes through the point (0,0). is concave down until it hits the point (1, -2), where it hits an inflection point.
The graph is then concave up, hitting a minimum at the point (2,-4), then increases at an increasing rate through the point (0,3).
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F ' (x) = f(x) means that F(x) is an antiderivative of f(x).
So F(x) will have critical points where f(x) = 0, F(x) will be increasing where f(x) is positive, decreasing where f(x) is negative, etc..
You appear to have analyzed the function f(x) rather than F(x).
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7) *6.1.12 Sketch and describe two functions F with F'(x) = f(x) (Sketch or graph f(x) with a graphing calculator) by only looking at the graph of f(x) = x^3 - 4x^2 + x + 6. In one let F(0) = 0 and the other F(0) = 1. Also identify local minima, maxima, and points of inflection.
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f'(x) = 3x^2 -8x + 1,
slope = 0, either minimum or maximum at f'(x) = 0
0 = 3x^2 - 8x + 1
x = 8 +- 'sqrt((-8^2) - 4 * 3 * 1) / (2*3) = (8 +- 'sqrt (64 - 12)) / 6 = (8 +- 'sqrt (52)) / 6
x = approx 2.53
x = approx 0.13
F(0.13) = 6.06, max at point approx ( 0.13, 6.06)
F(2.53) = -0.879, min at point approx(2.53, -0.879)
F''x = 6x - 8
Inflection at f''x = 0
x = 4/3 = 1.25
f(1.25) = 2.95
inflection at (1.25, 2.95)
The graph at f(0)= 1
it increases at decreasing rate until it hit a max at point ( 0.13,6.06) and is concave down.
The graph hits an inflection point at the point (1.25, 2.95) and becomes concave up.
the graph decreases at a decreasing rate and hits a minimum at the point (2.53, -0.879)
From the last the graph then increases at an increasing rate.
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8) *6.1.18 Let g' ={-x when 0 <= x <= 10, 4x when 10 <= x <= 15, x when 15 <= x <= 20, and -(1/2)x when x >= 20}. Sketch g' and given g(0) = 50 sketch the graph of g(x) and specify all critical and inflection points of g(x).
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g'x is a series of straight lines either increasing or decreasing.
g'(x) starts at the point (0,0) decreases at at a constant rate to the point (10,-10)
g'(x) increases at a constant rate from the point ( 10,-10) to (15, 50). it crosses the f'x axis at the point (12.5,0)
g'(x) decreases at a constant rate from (15,50) through ( 20, 40) and continues to decrease at a constant rate.
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g(x) is decreasing where g ' (x), the function which represents the behavior of its slope, is negative, and is decreasing at a decreasing rate where g ' (x) is negative and decreasing.
So g(x) decreases at a decreasing rate from x = 0 to x = 10.
At that point the slope becomes positive (and quite steep with an initial slope of 40) and increases at an increasing rate until x = 15.
The slope suddenly changes at x = 15.
etc..
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Area g'(x) from x = 0 to x = 10
= 1/2 * (-10 * 10) = -50
g(x)goes from the point (0,50) to ( 10,0)
g(x) continues to decrease at a decreasing rate until it hits the point (12.5, -12.5) where it turns and increases at an increasing rate.
g(x) then rises to the point (15, 62.5) where it hits an iflection point and becomes concave down.
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Between x = 10 and x = 15 the average value of g ' (x) is 50. Its integral on that interval is 200.
g ' remains positive from x = 15 to x = 20.
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g(x) increases to a point of (20, 87.5)
9) 6.1.22 Use a graph of 4cos(x^2) to determine where an antiderivative, F, of this function reaches its maximum on 0<= x<= 3. If F(1) =4, find this maximum obtained.
The graph of f'x has a slope a value of approximately 4.
This causes the graph of F(x) to take a max value at F(3) = 12.
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F has critical points where its derivative is zero.
cos(theta) = 0 when theta = pi/2 or 3 pi/2 or 5 pi/2 etc..
So if x^2 takes one of these values there is a critical point.
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10) Given the graph below, which depicts outflow vs. clock time and inflow vs. clock time for a reservoir:
When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves.
When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.
*** See the text section for this graph. In the current edition the problem is number 25; in other editions it shouldn't be far from that. ***
max when Inflow > outflow by max point.
&#Good responses. See my notes and let me know if you have questions. &#