#$&* course Mth 173 6/23 23:00 Calculus IIAsst # 1
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18:28:51
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Optional preliminary questions: ********************************************* Question: `q001. The domain of f(x) is the interval 1 <= x <= 5. Its graph is a straight line from (1, 3) to (5, 5). Sketch the graph of f(x), its derivative function f ' (x), and the function F(x) whose value at x = 1 is 7 and whose derivative F ' (x) is equal to f(x). Answer the following without symbolically integrating or differentiating f(x): Describe the graph of its derivative function f ' (x) on this interval. Describe the graph of its antiderivative function F(x), with the initial condition F(1) = 7. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of f'(x) is a horizontal straight line starting at the point (1, 0.5) to ( 5, 0.5) The graph of the AntiDerivative is a line that increases at a constant rate of 1 unit of rise per 2 units of run. It starts at the point (1,7) and ends at the point (5,9) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q002. The domain of g(x) is the interval 1 <= x <= 8. Its graph consists of a straight line from (1, 3) to (5, 5) (so that it coincides with the graph of the function f(x) of the preceding problem on the interval for which f(x) is defined) and then along another straight line from (5, 5) to (8, 12). Sketch the graph of g(x), its derivative function g ' (x), and the function G(x) whose value at x = 1 is 7 and whose derivative G ' (x) is equal to g(x). Describe the graph of its derivative function g ' (x) on this interval. Describe the graph of its antiderivative function G(x), with the initial condition G(1) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of f'(x) starts at the point (1,0.5) and is a horizontal line to (5, 0.5) the line then jumps up to f'(x) = 7/3, and is a straight line from x= 5 to x= 8, between the points (5, 7/3) to (8, 7/3) The graph with G(1) = 2 starts at the point (1,2) increases at a constant rate to the point( 4,5) then increases again at a costant rate to the point (12,8). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q003. The domain of h(x) is 0 <= x <= 6. Its graph consists of a smooth curve passing through the points (1, 3), (4, 5) and (6, 1), with its peak at (4, 5). The graph increases at a decreasing rate between x = 1 and x = 4, and decreases at an increasing rate between (4, 5) and (6, 1). Sketch the graph of h(x), its derivative function h ' (x), and the function H(x) whose value at x = 1 is 7 and whose derivative H ' (x) is equal to h(x). Based on your graph of h(x), what is your best estimate for the maximum value of h ' (x) on the interval 0 <= x <= 6? What is your best estimate for the minimum value of h ' (x) on this interval? What are the coordinates of the point at which h ' (x) crosses the x axis, if indeed it does so? Based on your graph of h(x), what is your best estimate for the change in the value of H(x) between x = 1 and x = 4, and for the change in the value of H(x) between x = 4 and x = 6? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Maximum value of h'(x) = 3, and occurs between the interval 0<=x<=1. Minimum value of h'(x) = -4, and occurs at x = 6. h'(x) crosses the x axis at about (0,4) Best estimate of change in H(x) x=1 to x= 4 'dy/'dx = (8.5 - 7) / ( 4-1) = 1.5 / 3 = 1/2 Best estimate of change in H(x) x=4 to x =6 , 'dy/'ds = (6.5 - 8.5) / (6 -4) = -2 / 2 = -1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:
#$&* course Mth 174 6/23 23:00 Had the wrong course number last submission, everything else copy and pasted. Calculus II
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18:28:51
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Optional preliminary questions: ********************************************* Question: `q001. The domain of f(x) is the interval 1 <= x <= 5. Its graph is a straight line from (1, 3) to (5, 5). Sketch the graph of f(x), its derivative function f ' (x), and the function F(x) whose value at x = 1 is 7 and whose derivative F ' (x) is equal to f(x). Answer the following without symbolically integrating or differentiating f(x): Describe the graph of its derivative function f ' (x) on this interval.
#$&* course Mth 174 6/23 23:00 Had the wrong course number last submission, everything else copy and pasted. Calculus II
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18:28:51
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Optional preliminary questions: ********************************************* Question: `q001. The domain of f(x) is the interval 1 <= x <= 5. Its graph is a straight line from (1, 3) to (5, 5). Sketch the graph of f(x), its derivative function f ' (x), and the function F(x) whose value at x = 1 is 7 and whose derivative F ' (x) is equal to f(x). Answer the following without symbolically integrating or differentiating f(x): Describe the graph of its derivative function f ' (x) on this interval. Describe the graph of its antiderivative function F(x), with the initial condition F(1) = 7. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of f'(x) is a horizontal straight line starting at the point (1, 0.5) to ( 5, 0.5)
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Given Solution: the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1. If f(3) = 0 then f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1. Working back from x=3, f(2) = 0 - (-1) = 1 and f(0) = 1 - 2 = -1. The integral is the sum of the changes in f ' which is 2 - 1 + 2 - 4 + 1 = 0. Alternatively since f(0) = -1 and f(7) = -1 the integral is the difference f(7) - f(0) = -1 - (-1) = 0. Let me know if you disagree with or don't understand any of this and I will explain further. Let me know specifically what you do and don't understand. ** ** Alternative solution: Two principles will solve this problem for you: 1. The definite integral of f' between two points gives you the change in f between those points. 2. The definite integral of f' between two points is represented by the area beneath the graph of f' between the two points, provided area is understood as positive when the graph is above the x axis and negative when the graph is below. We apply these two principles to determine the change in f over each of the given intervals. Answer the following questions: What is the area beneath the graph of f' between x = 0 and x = 2? What is the area beneath the graph of f' between x = 3 and x = 4? What is the area beneath the graph of f' between x = 4 and x = 6? What is the area beneath the graph of f' between x = 6 and x = 7? What is the change in the value of f between x = 3 and x = 4? Since f(3) = 0, what therefore is the value of f at x = 4? Now that you know the value of f at x = 4, what is the change in f between x = 4 and x = 6, and what therefore is the value of f at x = 6? Using similar reasoning, what is the value of f at x = 7? Then using similar reasoning, see if you can determine the value of f at x = 2 and at x = 0.** STUDENT QUESTION: I did not understand how to obtain the value of f(0), but I found that f(7) was 10 by adding all the integrals together INSTRUCTOR RESPONSE: The total area is indeed 10, so you're very nearly correct; however the integral is like a 'signed' area--areas beneath the x axis make negative contributions to the integral--and you added the 'absolute' areas &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: (Note that no problem reference is given, meaning that this question applies to the current problem. Any question that is not preceded by a problem number is likely to be in reference to the current problem.) 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) starts at the point (0,-1) and increases at an constant rate to the point (2,1). f(x) then decreases at a constant rate to the point (3,0) f(x) then increases at a constant rate to the point (4,2) f(x) then then decreases at a constant rate to the point (6,-2) f(x) then increases at a constant rate to the point (7,-1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph of f(x) is increasing, with slope 1, on the interval (0,2), since f'(x) = 1 on that interval, decreasing, with slope -1, on the interval (2,3), where f'(x) = -1, increasing, with slope +2, on the interval (3,4), where f'(x) = +2, decreasing, with slope -2, on the interval (4,6), where f'(x) = -2, and increasing, with slope +1, on the interval (6,7), where f'(x) = +1. The concavity on every interval is zero, since the slope is constant on every interval. Since f(3) = 0, f(4) = 2 (slope 2 from x=3 to x=4), f(6) = -2 (slope -2 from x = 4 to x = 6), f(7) = -1 (slope +1 from x=6 to x=7). Also, since slope is -1 from x=2 to x=3, f(2) = +1; and similar reasoning shows that f(0) = -1. ** ** The definite integral of f'(x) from x=0 to x=7 is therefore f(7) - f(0) = -1 - (-1) = 0. ** ** Basic principles: 1. The slope of the graph of f(x) is f'(x). So the slope of your f graph will be the value taken by your f' graph. 2. Note that if the slope of the f graph is constant for an interval that means that the graph is a straight line on the interval. Using these principles answer the following questions: What is the slope of the f graph between x = 0 and x = 2? What is the slope of the f graph between x = 3 and x = 4? What is the slope of the f graph between x = 4 and x = 6? What is the slope of the f graph between x = 6 and x = 7? Given that f(3) = 0 and using the value of the slope of the f graph between x = 3 and x = 4, describe the f graph between these two points. Using similar information describe the graph for each of the other given intervals. Also answer the following: What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): What is the slope of the f graph between x = 0 and x = 2? = 1 What is the slope of the f graph between x = 3 and x = 4? = 2 What is the slope of the f graph between x = 4 and x = 6? = -2 What is the slope of the f graph between x = 6 and x = 7? = 1 Given that f(3) = 0 and using the value of the slope of the f graph between x = 3 and x = 4, describe the f graph between these two points. Graph between these points has a slope of +2 and increases 2 values of y for 1 unit of x. Using similar information describe the graph for each of the other given intervals. Interval of x ( 0,2) slope value by +1, and goes from (0,-1) to (2,1) ' Interval of x (2,3) slope value = -1 interval of x ( 4,6) slope value = -2 intervl of x (6,7) slope value = +1 Also answer the following: What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. ** for f to be concave up, f' needs to take values below y = 0 on the f'(x) graph for f to be concave down, 'f' need to take values above y = 0 on the f'(x) graph anywhere it crosses the f'x = 0 axis woud be a inflection point. ------------------------------------------------ Self-critique Rating: ********************************************* Question: Was the graph of f(x) continuous? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes the graph of f(x) is continous it has no double values for any points of x. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a). Is this condition fulfilled at every point of the f(x) graph? **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: How can the graph of f(x) be continuous when the graph of f ' (x) is not continuous?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f'x has sudden turns of slope, but this does not cause the graph of f(x) to have any discontinities. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 18:38:11
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Given Solution: ** If f ' is constant then the slope of the f(x) graph is constant, so the graph of f(x) must be linear ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Section 6.1, Problem 10: The graph of outflow vs. time is concave up Jan 1993 -Sept, peaks ub October, then decreases somewhat thru Jan 1994; the inflow starts lower than the outflow, peaks in May, then decreases until January; inflow is equal to outflow around the middle of March and again in late July. **** When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Quanity of water is greatest when inflow - outflow = max value. This happens May 2007. Quantity of water is least when outflow - inflow = max value this happens October 2007. It looks like the max water level in the reservoir when inflow of water is at the highest absolute value over the inflow of water. its vice versa for the min water level. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Between any two dates the corresponding outflow is represented by the area under the outflow curve, and the inflow by the area under the inflow curve. When inflow is greater than outflow the quantity of water in the reservoir will be increasing and when the outflow is greater than the inflow quantity of water will be decreasing. We see that the quantity is therefore decreasing from January 93 through sometime in late February, increasing from late February through the beginning of July, then again decreasing through the end of the year. The reservoir will reach a relative maximum at the beginning of July, when the outflow rate overtakes the inflow rate. The amount of water lost between January and late February is represented by the difference between the area under the outflow curve and the area under the inflow curve. This area corresponds to the area between the two graphs. The amount of water gained between late February and early July is similarly represented by the area between the two curves. The latter area is clearly greater than the former, so the quantity of water in the reservoir will be greater in early July than on Jan 1. The loss between July 93 and Jan 94, represented by the area between the two graphs over this period, is greater than the gain between late February and early July, so the minimum quantity will occur in Jan 94. The rate at which the water quantity is changing is the difference between outflow and inflow rates. Specifically the net rate at which water quantity is changing is net rate = inflow rate - outflow rate. This quantity is represented by the difference between the vertical coordinate so the graphs, and is maximized around late April or early May, when the inflow rate most greatly exceeds the outflow rate. The net rate is minimized around early October, when the outflow rate most greatly exceeds the inflow rate. At this point the rate of decrease will be maximized. ** ** When inflow is > outflow the amount of water in the reservoir will be increasing. If outflow is < inflow the amount of water will be decreasing. Over what time interval(s) is the amount of water increasing? March 2007 thru July 2007 Over time interval(s) is the amount of water decreasing? ** Jan 2007 thru March 2007 and July 2007 thru Jan 2008
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**** When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.
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18:47:03 The curve increases most between Jan and Apr and it decreases most between July and October ** What aspect of which graph gives you the rate at which water is flowing into the reservoir? The graph of inflow gives me the rate at which water is flowing into the reservoir. What aspect of which graph gives you the rate at which water is flowing out of the reservoir? The graph of out flow gives me the rate at which water is flowing out of the reservoir. What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at an increasing rate? Only when the difference with inflow > outflow is positive and increasing for the water to be increasing at an increasing rate. What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at a decreasing rate? Only when inflow > outflow and the difference between the two is decreasing with time. What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at an increasing rate? Only when outflow > inflow, and the difference between the two is increasing. What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at a decreasing rate? ** Only when outfow > inflow, and the difference between the two is decreasing.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: Section 6.2, Problem 5 [[6.2.5 (previously 6.2 #26)]] antiderivative of f(x) = x^2, F(0) = 0
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: problem 6.2, # 33, edition 5 intergral ( x^2, 'dx, f(b), 0 ) = (1/3) * x^3 = x^3 / 3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** An antiderivative of x^2 is x^3/3. The general antiderivative of x^2 is F(x) = x^3/3 + c, where c can be anything. There are infinitely many possible specific antiderivative. However only one of them satisfied F(0) = 0. We have F(0) = 0 so 0^3/3 + c = 0, or just c = 0. The antiderivative that satisfies the conditions of this problem is therefore F(x) = x^3/3 + 0, or just F(x) = x^3/3. **
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18:47:58 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: Section 6.2, Problem 8 [[(previously 6.2 #56)]] indef integral of t `sqrt(t) + 1 / (t `sqrt(t)) **** What did you get for the indefinite integral? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Section 6,2 edition 5 # 51. indef integral of t `sqrt(t) + 1 / (t `sqrt(t)) = intergral ( t^(3/2) = t^(-3/2), 'dt, b,a) = 5/2 t ^(5/2) - 2/sqrt(t) + C confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The function can be written t^(3/2) + t^(-3/2). Both are power functions of the form t^n. Antiderivative is 2/5 * t^(5/2) - 2 t^(-1/2) + c or 2/5 t^(5/2) - 2 / `sqrt(t) + c. **
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11:39:51 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: 6.2.9 (previously 6.2 #50) definite integral of sin(t) + cos(t), 0 to `pi/4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = - cost |(pi/4 - 0) + sint | (pi/4 - 0) = (-1/sqrt'(2) + 1) + (1/sqrt(2) + 0) = 1 + C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** An antiderivative is -cos(t) + sin(t), as you can see by taking the derivative. Evaluating this expression at `pi/4 gives -`sqrt(2)/2 + `sqrt(2)/2 = 0. Evaluating at 0 gives -1 + 0 or -1. The antiderivative is therefore 0 - (-1) = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: Why doesn't it matter which antiderivative you use?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because if you take the derivtive of your this antidervative you get back the original function no matter what the constant C. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** General antiderivative is -cos(t) + sin(t) + c, where c can be any number. You would probably use c = 0, but you could use any fixed value of c. Since c is the same at both limits of the integral, it subtracts out and has no effect on the value of the definite integral. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: 6.2.13 (previously 6.2 #60) The average of v(x) = 6/x^2 on the interval [1,c} is 1. Find the value of c.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: intergral (6*x^-2,dx,c,1) = -6/x |c,1 = -6/c - -6/1= -6/c + 6 Definate intergral = averageValue * length of interval = 1 * (c - 1) = c-1 -6/c + 6 = c - 1 -6/c + 7 = c c + 6/c - 7 = 0 c*(c + 6/c - 7) = 0 * c c^2 + 6 - 7c = 0 c^2 - 7c + 6 = 0 (c - 6) *(c-1) = 0 c = 6 or c = 1 if c = 1, then the length of the interval = c -1 =1-1 = 0 = not practical. if c = 6, then the length of the intercal = c - 1 = 6-1 = 5 c = 6 Area under is equal to intergral (6x^-2, dx, 6, 1) = -6/x + 6 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: An antiderivative of 6 / x^2 is F(x) = -6 / x. The definite integral is equal to the product of the average value and the length of the interval. In this case average value is 1 and the interval from x = 1 to x = c has length c - 1. So the definite integral must be 1 * ( c - 1). Evaluating between 1 and c and using the above fact that the result must be 1 we get F(c) - F(1) = -6/c- (-6/1) = c - 1 so that -6/c+6=c - 1. We solve for c, first getting all terms on one side: c - 7 + 6/c = 0. Multiplying both sides by c to get c^2 - 7 c + 6 = 0. Either be factoring or the quadratic formula we get c = 6 or c = 1. If c = 1 the interval has length 0 and the definite integral is not defined. This leaves the solution c= 6. STUDENT QUESTION I had some trouble with this problem I got -6/x for antideri. So I thought that at F(1) = -6 and F(1.5) = -4 Then I got really confused for some reason used the logic F(b)- F(a) = -4 - (-6) = 2 when divided by 2 = 1. I see what you did but not so sure about the logic. INSTRUCTOR RESPONSE Note that the length of the interval between x = 1 and x = 1.5 is .5. The integral is 2, but the average value between x=1 and x=1.5 is (integral) / (length of interval) = 2 / .5 = 4, not 2. The average value of the integral must be 1. The integral of a function over an interval is equal to its average value over that interval, multiplied by the length of the interval: ave value = definite integral / length of interval It follows immediately that definite integral = ave value * length of interval In this case the interval has length (c - 1) and the average value must be 1. The integral must therefore be 1 * (c - 1). The integral is from x = 1 to x = c. So The integral of 6/x^2 from x = 1 to x = c must equal 1 * (c - 1). The integral of 6/x^2 from x = 1 to x = c is -6 / c - (-6 / 1) = 6 - 6/c. Thus 6 - 6/c = 1 * (c - 1). We solve to get c, and we obtain c = 6.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: 6.2.14 (previously 6.2 #44) What is the indefinite integral of e^(5+x) + e^(5x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: indefinite integral of e^(5+x) + e^(5x) = intergral (e^(5+x), dx) + intergral ( e^5x, dx) = e^u * 'du = e^u 'du = e^(5+x) = e^(5+x) + C + intergral ( e^(5x) dx) = 1/5 *e^(5x) + c = e^(5+x) + 1/5 * e^5x + C confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The derivative of e^(5+x) is, by the Chain Rule, (5+x)' * e^(5+x) = 1 * e^(5 + x) = e^(5 + x) so this function is its own antiderivative. The derivative of e^(5x) is (5x) ' * e^(5x) = 5 * e^(5x). So to get an antiderivative of e^(5x) you would have to use 1/5 e^(5x), whose derivative is e^(5x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. The domain of g(x) is the interval 0 <= x <= 4. Its graph consists of a straight line from (0, 3) to (2, 6) and then along a curve which increases at a decreasing rate from (2, 6) to (4, 8), reaching slope zero at that final point. Sketch the graph of g(x), its derivative function g ' (x), and the function G(x) whose value at x = 1 is 7 and whose derivative G ' (x) is equal to g(x). Describe the graph of its derivative function g ' (x) on this interval. Describe the graph of its antiderivative function G(x), with the initial condition G(0) = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g'(x) interval 0<=x<2 = 'dy / 'dx = ( 6-3) / (2 -0) = 3/2 = 1.5 the graph of g'(x) on the interval (0,2) is a straight line at g'(x) = 1.5. The graph of g'(x) then decreases at an increasing rate through the point (4,0) where the slope = 0. The graph of the anti deritive funtion is a straight line on the interval (0,2) from (0,0) to (2,3)