course MTH 271 Ѫv`assignment #001
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13:58:49 **** Query problem 1.4.06 diff quotient for x^2-x+1 **** What is the simplified form of the difference quotient for x^2-x+1?
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14:03:17 ** The difference quotient would be [ f(x+`dx) - f(x) ] / `dx = [ (x+`dx)^2 - (x+`dx) + 1 - (x^2 - x + 1) ] / `dx. Expanding the squared term, etc., this is [ x^2 + 2 x `dx + `dx^2 - x - `dx + 1 - x^2 + x - 1 ] / `dx, which simplifies further to }[ 2 x `dx - `dx + `dx^2 ] / `dx, then dividing by the `dx we get 2 x - 1 + `dx. For x = 2 this simplifies to 2 * 2 - 1 + `dx = 3 + `dx. **
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14:24:57 problem 1.4.34f+g, f*g, f/g, f(g), g(f) for f=x/(x+1) and g=x^3 the requested functions and the domain and range of each.
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RESPONSE --> f+g= x/(x+1) + x^3=(x+ x^4+ x^3)/x+1 domain= -inf to -1 to inf range =-inf to -1 to inf f*g= x/(x+1) * x^3= x^4/(x+1) domain -inf to -1 to inf range -inf to -1 to inf f/g= x/(x+1)/x^3= x/(x^4+x^3) domain -inf to 0 to inf range -1,1 except at 0
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14:25:58 ** (f+g)(x) = x / (x + 1) + x^3 = (x^4 + x^3 + x) / (x + 1). Domain: x can be any real number except -1. (f * g)(x) = x^3 * x / (x+1) = x^4 / (x+1). Domain: x can be any real number except -1. (f / g)(x) = [ x / (x+1) ] / x^3 = 1 / [x^2(x+1)] = 1 / (x^3 + x), Domain: x can be any real number except -1 or 0 f(g(x)) = g(x) / (g(x) + 1) = x^3 / (x^3 + 1). Domain: x can be any real number except -1 g(f(x)) = (f(x))^3 = (x / (x+1) )^3 = x^3 / (x+1)^3. Domain: x can be any real number except -1 **
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14:27:19 problem 1.4.60 graphs of |x|+3, -.5|x|, |x-2|, |x+1|-1, 2|x| from graph of |x|
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14:28:09 ** The graph of y = | x | exists in quadrants 1 and 2 and has a 'v' shape with the point of the v at the origin. It follows that: The graph of | x |+3, which is shifted 3 units vertically from that of | x |, has a 'v' shape with the point of the v at (0,3). The graph of -.5 | x |, which is stretched by factor -.5 relative to that of | x |, has an inverted 'v' shape with the point of the v at (0,0), with the 'v' extending downward and having half the (negative) slope of the graph of | x |. The graph of | x-2 |, which is shifted 2 units horizontally from that of |x |, has a 'v' shape with the point of the v at (2, 0). The graph of | x+1 |-1, which is shifted -1 unit vertically and -1 unit horizontally from that of | x |, has a 'v' shape with the point of the v at (-1, -1). The graph of 2 |x |, which is stretched by factor 2 relative to that of | x |, has a 'v' shape with the point of the v at (0,0), with the 'v' extending upward and having double slope of the graph of | x |. |x-2| shifts by +2 units because x has to be 2 greater to give you the same results for |x-2| as you got for |x|. This also makes sense because if you make a table of y vs. x you find that the y values for |x| must be shifted +2 units in the positive direction to get the y values for |x-2|; this occurs for the same reason given above For y = |x+1| - 1 the leftward 1-unit shift is because you need to use a lesser value of x to get the same thing for |x+1| that you got for |x|. The vertical -1 is because subtracting 1 shifts y downward by 1 unit **
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RESPONSE --> i didn't understand the question
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14:31:45 problem 1.4.64 find x(p) from p(x) = 14.75/(1+.01x)
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RESPONSE --> (1+.01x)=14.75 .01x=13.75 x=1375
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14:32:56 ** p = 14.75 / (1 + .01 x). Multiply both sides by 1 + .01 x to get (1 + .01 x) * p = 14.75. Divide both sides by p to get 1 + .01 x = 14.75 / p. Subtract 1 from both sides to get 1 x = 14.75 / p - 1. Multiply both sides by 100 to get = 1475 / p - 100. Put the right-hand side over common denominator p: = (1475 - 100 p) / p. If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
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RESPONSE --> I set the equation = to 0 instead of P.
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14:33:28 What is the x as a function of p, and how many units are sold when the price is $10?
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14:33:36 ** If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
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jx}ʟ assignment #010 ݺӯD Applied Calculus I 07-08-2006
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14:56:56 What is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.
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RESPONSE --> f(x)= (x+3)(x-4)(x-9)= x^3-4x+x^2-9x+3x-12+3x-27+x^2-9x-4x+36= x^3+4x^2-12x-3 range goes from pos to neg as domain approaches 0 from -1, range keeps going negative until domain goes to 1. range increases toward real axis as domain increases above 1 crosses real axis when domain is >2 and <3
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14:57:29 ** A polynomial with zeros at -3, 4 and 9 must have factors (x + 3), ( x - 4) and (x - 9), and so must contain (x+3) (x-4) (x-9) as factors. These factors can be multiplied by any constant. For example 8 (x+3) (x-4) (x-9), -2(x+3) (x-4) (x-9) and (x+3) (x-4) (x-9) / 1872 are all polynomizls with zeros at -3, 4 and 9. If -3, 4 and 9 are the only zeros then (x+3), (x-4) and (x-9) are the only possible linear factors. It is possible that the polynomial also has irreducible quadratic factors. For example x^2 + 3x + 10 has no zeros and is hence irreducible, so (x+3) (x-4) (x-9) (x^3 + 3x + 10) would also be a polynomial with its only zeros at -3, 4 and 9. The polynomial could have any number of irreducible quadratic factors. **
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RESPONSE --> i did not multiply by a constant
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15:06:01 problem 1.5.16 right-, left-hand limits and limit (sin fn)
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RESPONSE --> left hand limits sin fn = -inf to -.0000001= .9999999
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15:09:09 What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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RESPONSE --> right side -.0000000000017 to -1 left side -.00000000000017 to 1 -1 to1
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15:10:07 ** Imagine walking along the graph from the right. As you approach the limiting x value, your y 'altitude' gets steadier and steadier, approaching closer and closer what value? You don't care what the function actually does at the limiting value of x, just how it behaves as you approach that limiting value. The same thing happens if you walk along the graph from the left. What does you y value approach? Is this the value that you approach as you 'walk in' from the right, as well as from the left? If so then it's your limit. **
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RESPONSE --> the value appears to be approaching o
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15:10:33 problem 1.5.20 right-, left-hand limits and limit (discont at pt) What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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15:10:56 STUDENT RESPONSE: The three limits are a. 0 b. 2 c. no limit The limit doesn't exist because while the function approaches left and right-handed limits, those limite are different. INSTRUCTOR COMMENT: That is correct. ADVICE TO ALL STUDENTS: Remember that the limit of a function at a point depends only on what happens near that point. What happens at the point iself is irrelevant to the limiting behavior of the function as we approach that point.
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15:13:09 problem 1.5.26 lim of (x+4)^(1/3) as x -> 4 What is the desired limit and why?
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RESPONSE --> the function is o at -4
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15:14:42 ** The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2. **
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RESPONSE --> i didn't understand this. x->4 what the negative symbol was for?
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15:21:08 problem 1.5.38 lim of (x^3-1)/(x-1) as x -> 1
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RESPONSE --> if x- is numbers on the left side of 1??? if not, numbers would be approaching 1 from the right side. then using 2, =7 using 1.000001 = 3 using 1.000000001 = 3 limit appears to be 3
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15:21:33 ** As x -> 0 both numerator and denominator approach 0, so you can't tell just by plugging in numbers what the limit will be. If you factor x^3-1 into (x-1)(x^2+x+1) you can reduce the fraction to (x-1)(x^2+x+1) / (x-1) = x^2 + x + 1, which is equal to the original function for all x except 1 (we can't reduce for x = 1 because x-1 would be zero, and we can't divide by zero). As x -> 1, x^2 + x + 1 -> 3. It doesn't matter at all what the function does at x = 1, because the limiting value of x never occurs when you take the limit--only x values approaching the limit count. 3 is therefore the correct limit. **
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15:25:03 problem 1.5.56 lim of 1000(1+r/40)^40 as r -> 6%
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RESPONSE --> 1000(1+.06/40)^40=1061.79
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15:25:17 ** $1000 *( 1+.06 / 40)^40 = 1061.788812. Since this function changes smoothly as you move through r = .06--i.e., since the function is continuous at r = .06--this value will be the limit. **
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15:25:23 Add comments on any surprises or insights you experienced as a result of this assignment.
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Lmѩʰ_뗄~ assignment #011 ݺӯD Applied Calculus I 07-08-2006
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15:27:11 problem 1.6.14 intervals of cont for (x-3)/(x^2-9) What are the intervals of continuity for the given function?
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RESPONSE --> 3, -3 graph is undefined
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15:27:43 ** The function is undefined where x^2 - 9 = 0, since division by zero is undefined. x^2 - 9 = 0 when x^2 = 9, i.e., when x = +-3. So the function is continuous on the intervals (-infinity, -3), (-3, 3) and (3, infinity). The expression (x - 3) / (x^2 - 9) can be simplified. Factoring the denominator we get (x - 3) / [ (x - 3) ( x + 3) ] = 1 / (x + 3). This 'removes' the discontinuity at x = +3. However in the given fom (x-3) / (x^2 + 9) there is a discontinuity at x = -3. **
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15:28:59 problem 1.6.22 intervals of cont for |x-2|+3, x<0; x+5, x>=0 What are the intervals of continuity for the given function?
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15:29:25 ** The graph of y = |x-2|+3 is translated 2 units in the x direction and 3 in the y direction from the graph of y = |x|. It forms a V with vertex at (2, 3). The given function follows this graph up to x = 0. It has slope -1 and y-intercept at y = | 0 - 2 | + 3 = 5. The graph then follows y = x + 5 for all positive x. y = x + 5 has y-intercept at y = 5. From that point the graph increases along a straight line with slope 1. So the graph of the given function also forms a V with vertex at (0, 5). Both functions are continuous up to that point, and both continuously approach that point. So the function is everywhere continuous. **
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15:31:13 problem 1.6.54 lin model of revenue for franchise Is your model continuous? Is actual revenue continuous?
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RESPONSE --> non continuous since it is per year. the revenue would be cont.
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