#$&* course Mth 158 12/6 11 036. * 36
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Given Solution: The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). The graph of y = 2^(x+2), being shifted in only the x direction, is also asymptotic to the negative x axis and passes through the points (0 - 2, 1) = (2, 1) and (1 - 2, 2) = (3, 2). The graph also increases at a rapidly increasing rate. All the points of the graph of y = 2^(x + 2) lie 2 units to the left of points on the graph of y = 2^x. To understand why the graph shifts to the left, and why we used the points (0 - 2, 1) = (2, 1) and (1 - 2, 2) = (3, 2) as a basis for the graph, consider the tables for y = 2^x and y = 2^(x + 2). The tables are given below: x y = 2^x x y = 2^(x+2) -3 1/8 -3 1/2 -2 1/4 -2 1 -1 1/2 -1 2 0 1 0 4 1 2 1 8 2 4 2 16 3 8 3 32 Observe that the y values 1/2, 1, 2, 4 and 8 in the y = 2^(x+2) column also occur in the y = 2^x column, but for different values of x: The values of x for the y = 2^(x+2) function, corresponding to y values 1/2, 1, 2 and 4, are 2 units less than for the y = 2^x function. This occurs because the exponent x + 2 is 2 units greater than the exponent x, so that x + 2 is always 2 units 'ahead' of the value of x. Thus y = 2^(x + 2) reaches its values 'earlier' than y = 2^x (for example y = 2^(x + 1) reaches the value y = 8 when x = 1, whereas y = 2^x doesn't reach y = 8 until x = 3). This causes the y = 2^(x + 2) graph to be shifted 2 units to the left, relative to the graph of y = 2^x. The figure below depicts the graphs of the two functions: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * Extra Problem / 7th edition 5.3.42. Transformations to graph f(x) = 1 - 3 * 2^x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started by making x:y charts with x= -3 through 3 First I started with basic y=2^x and got the coordinates (-3, .125); (-2, .25); (-1, .5); (0, 1); (1, 2); (2, 4); and (3,8). You already know what this looks like on a graph due to the previous question, but it increases to the right at an increasing rate. Then when I make a chart with x:y=-3*2^x I multiplied all of the y coordinates from the first chart by -3 so that the points become (-3, -.375); (-2, -.75); (-1, -1.5); (0, -3); (1, -6); (2, -12); and (3,-24) When you graph these coordinates, you get a slightly extended version of the line that decreases and an increasing rate but when I made a chart that was x:y=1-3*2^x I had to add 1 to all of the y coordinates which resulted in (-3, .625); (-2, .25); (-1, -.5); (0, -2); (1, -5); (2, -11); and (3,-23) and when you graph it, you find that the line just shifted up 1 unit confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). -3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate. To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate. If you don't understand the above, then do as follows, without looking up at the solution given so far: Plot the basic points (0, 1) and (1, 2) of the y = 2^x function. Multiply your y values by 3 to get the basic points of the y = 3 * 2^x function, plot your points and sketch the graph (this is your 'vertical stretch'; you should see that it moves your original points 3 times as far from the x axis as before. The same thing happens to all the points of the original y = 2^x graph--they all move 3 times as far from the x axis.) Multiply your y values by -1. This gives you the basic points of the y = - 3 * 2^x function, plot your points and sketch the graph.. Add 1 to your y values, plot your points and sketch your graph. This gives you the basic points of the y = 1 -3 * 2^x function. (It should be clear that this 'shifts' the points of your y = - 3 * 2^x graph 1 unit in the vertical direction). Having done this, look again at the given solution. You might also consider the following table: x y = 2^x y = 3 * 2^x y = -3 * 2^x y = 1 -3 * 2^x 0 1 3 -3 -2 1 2 6 -6 -5 It should be clear how this table demonstrates the process described above (get y values of basic function, multiply by 3, multiply by -1, add 1), and how you transform the 'basic points' from (0, 1) and (1, 2) to get (0, 3) and (1, 6), then (0, -3) and (1, -6) and finally (0, -2) and (1, -5). You should identify these points on the graph depicted below, and having idenfied the basic points you should be able to identify which function is which. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * Extra Problem / 7th edition 5.3.60 Solve (1/2)^(1-x) = 4. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I didn't understand the ln thing, but found it on my calculator. So I guess I sort of get that if you go through the normal process of solving, just with ln involved you get ln( (1/2)^(1 - x) = ln(4) and divide both sides by ln(1/2) (1 - x) = ln(4) / ln(1/2) and then plug ln(4) / ln(1/2) into the calculator to get -2 so (1-x)=-2 and then get x=3(not 1 xp) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The reliable way to solve a problem of this nature is to recognize that the variable x occurs in an exponent, so we can 'get at it' by taking logs of both sides: (1/2)^(1-x) = 4. Taking the natural log of both sides, using the laws of logarithms, we get ln( (1/2)^(1 - x) = ln(4)) so that (1 - x) ln(1/2) = ln(4) and (1 - x) = ln(4) / ln(1/2). A calculator will reveal that ln(4) / ln(1/2) = -2, so that 1 - x = -2. We easily solve for x, obtaining x = 1. It is also possible to reason this problem out directly, and in this case our reasoning leads us to an exact solution: We first recognize one fact: 4 is an integer power of 2, and 1/2 is an integer power of 2, so 4 must also be an integer power of 1/2. Since 4 = 2^2, and since 1/2 = 2^-1, we can recognize that 4 = 2^-2, and reason as follows: (1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2 If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 6.3.98 / 7th edition 5.3.78 A(n) = A0 e^(-.35 n), area of wound after n days. What is the area after 3 days and what is the area after 10 days? Init area is 100 mm^2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: plug 100 e^(-.35*3) in the calculator and approximately get 35 same thing with 100e^(-.35*10) and get approximately 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2. Since e^0 = 1 this tells us that A0 = 100 mm^2. So the function is A(n) = 100 mm^2 * e^(-.35 n). To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx.. After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm^3 * .0302 = 3.02 mm^2 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 6.3.104 / 7th edition 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Statistics? How cruel... I get how you plug in 5 or 8 where the x is, and I know how to put e on my calculator, but I don't know how to put in the '!' should I have known that? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx.. The probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-4 / (8 !) = .030 approx.. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: * 6.3.104 / 7th edition 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Statistics? How cruel... I get how you plug in 5 or 8 where the x is, and I know how to put e on my calculator, but I don't know how to put in the '!' should I have known that? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx.. The probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-4 / (8 !) = .030 approx.. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!