course Mth 173 ???????€???assignment #007s?r??????????????Calculus I
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17:18:12 Query class notes #07Explain how we obtain the tangent line to a y = k x^3 function at a point on its graph, and explain why this tangent line gives a good approximation to the function near that point.
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RESPONSE --> To find the tangent line we find the derivative of a function. For this particular function (y=kx^3), the derivative is y' = 3kx^2. This tangent line is a good approximation because it has the same slope as the line at that exact instant.
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17:18:24 ** If we know that y=kx^3, as in the sandpile model, we can find the derivative as y = 3kx^2. This derivative will tell us the rate at which the volume changes with respect to the diameter of the pile. On a graph of the y = k x^3 curve the slope of the tangent line is equal to the derivative. Through the given point we can sketch a line with the calculated slope; this will be the tangent line. Knowing the slope and the change in x we easily find the corresponding rise of the tangent line, which is the approximate change in the y = k x^3 function. In short you use y' = 3 k x^2 to calculate the slope, which you combine with the change `dx in x to get a good estimate of the change `dy in y. **
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RESPONSE --> Ok.
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17:19:17 Query class notes #08What equation do we get from the statement 'the rate of temperature change is proportional to the difference between the temperature and the 20 degree room temperature'? What sort of graph do we get from this equation and why?
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RESPONSE --> We would get something like y = k(T-20). For this type function we would have a line that decreases toward the asymptote at 20 degrees. This line would decrease slower and slower.
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????????????assignment #007 s?r??????????????Calculus I 06-20-2007
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17:22:22 STUDENT SOLUTION AND INSTRUCTOR COMMENT: Would it be y = x-20 degrees., with y being the rate of temperature change and x being the temperature?You get a graph with a straight line and a slope of -20? INSTRUCTOR COMMENT: Not a bad attempt. However, you wouldn't use y to represent a rate, but rather dy /dt or y'. An in this example I'm going to use T for temperature, t for clock time. Read further. We need a graph of temperature vs. clock time, not rate of change of temperature vs. clock time. The difference between temperature and room temperature is indeed (T - 20). The rate of change of the temperature would be dT / dt. To say that these to our proportional is to say that dT / dt = k ( T - 20). To solve the situation we would need the proportionality constant k, just as with sandpiles and other examples you have encountered. Thus the relationship is dT / dt = k ( T - 20). Since dT / dt is the rate of change of T with respect to t, it tells you the slope of the graph of T vs. t. So the equation tells you that the slope of the graph is proportional to T - 20. Thus, for example, if T starts high, T - 20 will be a relatively large positive number. We might therefore expect k ( T - 20) to be a relatively large positive number, depending on what k is. For positive k this would give our graph a positive slope, and the temperature would move away from room temperature. If we are talking about something taken from the oven, this wouldn't happen--the temperature would move closer to room temperature. This could be accomplished using a negative value of k. As the temperature moves closer to room temperature, T - 20 becomes smaller, and the steepness of the graph will decrease--i.e., as temperature approaches room temperature, it will do so more and more slowly. So the graph approaches the T = 20 value more and more slowly, approaching as an asymptote. **
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RESPONSE --> Ok. I just didn't label the function as a rate function.
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17:24:07 Query Inverse Functions and Logarithms, Problem 7. Construct table for the squaring function f(x) = x^2, using x values between 0 and 2 with a step of .5. Reverse the columns of this table to form a partial table for the inverse function.
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RESPONSE --> Ok. Here is my table: Squaring function: (0,0) (.5,.25) (1,1) (1.5,2.25) (2,4) Inverse function: (0,0) (.25,.5) (1,1) (2.25,1.5) (4,2)
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17:24:13 STUDENT SOLUTION: We get the following ordered pairs: Table 1-- (0,0),(.5,.25),(1,1),(1.5,2.25),(2,4) Table2--(0,0),(.25,.5),(1,1),(2.25,1.5),(4,2).
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RESPONSE --> Ok
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17:24:36 Plot the points corresponding to the table of the squaring function, and plot the points corresponding to the table of its inverse. Sketch a smooth curve corresponding to each function. The diagonal line on the graph is the line y = x. Connect each point on the graph of the squaring function to the corresponding point on the graph of its inverse function. How are these pairs of points positioned with respect to the y = x line?
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RESPONSE --> Both lines are equally distant from the y=x line at each point.
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17:25:23 ** The segments connecting the graph points for function and for its inverse will cross the y = x line at a right angle, and the graph points for the function and for the inverse will lie and equal distances on either side of this line. The graph of the inverse is therefore a reflection of the graph of the original function through the line y = x. **
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RESPONSE --> Ok. I missed the reflection through the origin and the idea of the segment through the y=x line.
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17:25:34 **** 8. If we reversed the columns of the 'complete' table of the squaring function from 0 to 12, precisely what table would we get?
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RESPONSE --> We would get a 'complete' square root function from 0 to 144.
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17:26:08 ** We would get a table of the square root function with the first column running from 0 to 144, the second column consisting of the square roots of these numbers, which run from 0 to 12. **
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RESPONSE --> Ok.
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17:26:52 Sketch the graphs of the functions described by both tables. 9. If we could construct the 'complete' table of the squaring function from 0 to infinity, listing all possible positive numbers in the x column, then why would we be certain that every possible positive number would appear exactly one time in the second column?
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RESPONSE --> Because one number has one positive square root only, along with that fact that every real number would be on the table somewhere.
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17:27:41 ** The table you constructed had only some of the possible x and y values. A complete table, which couldn't actually be written down but can to an extent be imagined, would contain all possible x values. We could be sure because every number is the square of some other number. If the function was, for example, x / (x^2 + 1) there would be a great many positive numbers that wouldn't appear in the second column. But this is not the case for the squaring function. **
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RESPONSE --> Ok.
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17:27:51 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> At 4.31, the second column would be 4.31^2, or 18.5761.
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17:27:56 ** In the original table the second column would read 18.57, approx.. This is the square of 4.31. **
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RESPONSE --> Ok.
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17:28:01 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> At 'sqrt(18), the second column would be 18.
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17:28:05 ** 18 would appear in the second column because the square of sqrt(18) is 18. **
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RESPONSE --> Ok.
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17:28:15 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> At 'pi, the second number would be 'pi^2 or roughly 9.8696.
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17:28:19 ** The number would be `pi^2 **
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RESPONSE --> Ok.
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17:28:38 What would we obtain if we reversed the columns of this table?
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RESPONSE --> If we reversed the columns, we would get a 'complete' square root table from 0 to infinity.
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17:28:55 STUDENT ANSWER: We would obtain the inverse, the square roots of the squares being in the y colume and the squared numbers being in the x column.
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RESPONSE --> Ok. I didn't label the columns.
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17:29:03 What number would appear in the second column next to the number 4.31 in the first column of this table?
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RESPONSE --> At 4.31 in this table, the second column would be 'sqrt(4.31) = 2.076.
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00:04:36 This number would be 4.31 squared,18.5761.
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RESPONSE --> Shouldn't it be the square root of 4.31?
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00:05:20 What number would appear in the second column next to the number
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RESPONSE --> Hopefully this is the 'pi question. At 'pi^2, the second column would be 'pi.
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00:05:30 `pi^2 in the first column of this table?
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RESPONSE --> See prior response.
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00:05:34 STUDENT ANSWER: This number would be the square root, 'pi
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RESPONSE --> Ok.
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00:06:12 What number would appear in the second column next to the number -3 in the first column of this table?
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RESPONSE --> At -3, the second column would be nothing since you can't take the square root of a negative number.
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00:06:22 ** The number -3 doesn't appear in the second column of the original table so it won't appear in the first column of the inverted table. Note that sqrt(-3) is not a real number, since the square of a real number must be positive. **
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RESPONSE --> Ok.
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00:09:13 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18 2 ^ (4x) = 12 5 * 2^x = 52 2^(3x - 4) = 9.
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RESPONSE --> 2^x = 18: log(18)/log(2) = 4.1699 2^(4x) = 12: (log(12)/log(2)) / 4 = 0.89624 5*2^x = 52: log(52/5)/log(2) = 3.37851 2^(x3 - 4) = 9: ((log(9)/log(2)) + 4) / 3 = -.27669
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00:10:04 b^x = c is translated into logarithmic notation as log{base b}(c) = x. So: 2^x = 18 translates directly to log{bas3 2}(18) = x. For 5 * 2^x = 52, divide both sides by 5 to get 2^x = 10.4. Now take logs: 2x = log{base 2}(10.4) so x = 1/2 log{base 2}(10.4). Evaluate on your calculator. 2^(3x-4) = 9 translates to log{base 2}(9) = 3x - 4.
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RESPONSE --> Ok. I think I got that.
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00:10:29 14. Solve 2^(3x-5) + 4 = 0
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RESPONSE --> No solution since you can't take the log of a negative number.
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00:10:34 **
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RESPONSE --> Hm?
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00:10:44 2^(3x-5) + 4 = 0 rearranges to 2^(3x-5) =-4, which we translate as 3x-5 = log {base 2}(-4) = log(-4) / log (2). However log(-4) doesn't exist. When you invert the 10^x table you don't end up with any negative x values. So there is no solution to this problem. Be sure that you thoroughly understand the following rules: 10^x = b translates to x = log(b), where log is understood to be the base-10 log. e^x = b translates to x = ln(b), where ln is the natural log. a^x = b translates to x = log{base a} (b), where log{base a} would be written in your text as log with subscript a. log{base a}(b) = log(b) / log(a), where log is the base-10 log. It also works with the natural log: log{base a}(b) = ln(b) / ln(a).
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RESPONSE --> Ok.
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00:10:48 **
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RESPONSE --> ?
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00:11:04 Solve 2^(1/x) - 3 = 0
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RESPONSE --> x = 0.630929
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00:11:42 ** Rearrange to 2^(1/x) = 3. Then take log of both sides: log(2^(1/x) ) = log(3). Use properties of logs: (1/x) log(2) = log(3). Solve for x: x = log(2) / log(3). **
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RESPONSE --> Ah. I just didn't write down all the steps. I also didn't put it in log notation.
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00:12:50 Solve 2^x * 2^(1/x) = 15
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RESPONSE --> Hmm. Not quite sure.
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00:13:24 ** 2^x * 2^(1/x) = 15. By the laws of exponents we get 2^(x +1/x) = 15 so that x + 1/x = log {base2}(15) or x + 1/x =log(15) / log(2). Multiply both sides by x to get x^2 + 1 = [log(15) / log(2) ] * x. This is a quadratic equation. }Rearrange to get x^2 - [ log(15) / log(2) ] * x + 1 = 0 or x^2 - 3.91 * x + 1 = 0. Solve using the quadratic fomula. **
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RESPONSE --> Wow. I need to study that one more.
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00:15:19 Solve (2^x)^4 = 5
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RESPONSE --> First we have to reduce the power of 4: 2^x = 5^(1/4) Then take the standard base 2 log to find x: log(5^(1/4)) / log(2) = 0.580482.
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00:15:51 ** log( (2^x)^4 ) = log(5). Using laws of logarithms 4 log(2^x) = log(5) 4 * x log(2) = log(5) 4x = log(5) / log(2) etc.**
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RESPONSE --> Ok. Well, I think my way worked too.
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00:48:14 problem 1.3.10. C=f(A) = cost for A sq ft. What do f(10k) and f^-1(20k) represent?
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RESPONSE --> f(10k) represents a function that returns the cost of a 10k square foot building. f^-1(20k) is the inverse, a function that takes 20k (a cost) and turns it into the maximum square feet for a building.
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00:48:19 ** f(10,000) is the cost of 10,000 sq ft. f^-1(20,000) is the number of square feet you can cover for $20,000. **
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RESPONSE --> Ok.
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00:48:51 problem 1.3.20. vert stretch y = x^2 by factor 2 then vert shift 1.
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RESPONSE --> y = 2x^2 + 1.
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00:48:59 ** Vertically stretching y = x^2 we get y = 2 x^2. The vertical shift adds 1 to all y values, giving us the function y = 2 x^2 + 1. **
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RESPONSE --> Ok.
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00:49:11 Give the equation of the function.Describe your sketch in detail.
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RESPONSE --> Of what function? The previous one? Okay: y = 2x^2 + 1. This sketch is the typical squared function parabola which has been stretched by a factor of 2 and moved 1 up.
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00:50:37 ** The function would be y = f(x) = 2 x^2 + 1. The factor 2 stretches the y = x^2 parabola verticall and +1 shifts every point of the stretched parabola 1 unit higher. The result is a parabola which is concave up with vertex at point (0,1). The parabola has been stretched by a factor of 2 as compared to a x^2 parabola. If the transformations are reversed the the graph is shifted downward 1 unit then stretched vertically by factor 2. The vertex, for example, shifts to (0, -1) then when stretched shifts to (0, -2). The points (-1, 1) and (1, 1) shift to (-l, 1) and (1, 0) and the stretch leaves them there. The shift would transform y = x^2 to y = x^2 - 1. The subsequent stretch would then transform this function to y = 2 ( x^2 - 1) = 2 x^2 - 2. The reversed pair of transformations results in a parabola with its vertex at (0, -2), as opposed to (0, -1) for the original pair of transformations. **
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RESPONSE --> Hmm. Much more detailed than my description.
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00:54:24 problem 1.3.35 (was 1.8.30) estimate f(g(1))what is your estimate of f(g(1))?Explain how you look at the graphs of f and g to get this result
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RESPONSE --> I don't know which number this is. 1.3.35 is a question asking if a function was even, odd, or neither. (I got even). There are several like this in section 1.3: 14, 15, and 16. However, I still don't know which one it is.
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00:54:31 *&*& right problem? *&*&
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RESPONSE --> Hmm?
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00:55:17 ** You have to first find g(1), then apply f to that value. To find g(1), you note that this is g(x) for x = 1. So you look on the x-axis where x = 1. Then you move up or down to find the point on the graph where x = 1 and determine the corresponding y value. On this graph, the x = 1 point lies at about y = 2. Then you look at the graph of f(x). You are trying to find f(g(1)), which we now see is f(2). So we look at the x = 2 point on the x-axis and then look up or down until we find the graph, which for x = 2 lies just a little bit above the x axis. Looking over to the y-axis we see that at this point y is about .1. **
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RESPONSE --> Okay. Interesting.
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00:56:08 problem 1.5.12 graph, decide if inverse, approximate inverse at x = 20 for f(x) = x^2+e^x and g(x) = x^3 + 3^x
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RESPONSE --> This one was not assigned.
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00:56:25 ** The inverse of a function at a certain value is the x that would give you that value when plugged into the function. At x = 20 for g(x) = x^2 + e^x is the x value for which x^3 + 3^x = 20. The double use of x is confusing and way the problem is stated in the text isn't as clear as we might wish, but what you have to do is estimate the required value of x. It would be helpful to sketch the graph of the inverse function by reflecting the graph of the original function through the line y = x, or alternatively and equivalently by making an extensive table for the function, then reversing the columns. **
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RESPONSE --> Okay.
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00:58:00 problem 1.5.15 t(x) increasing over all reals with range (0,4) describe your possible graph of t^-1 (x), including a description of domain and range, increasing or decreasing behavior, and concavity
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RESPONSE --> The domain would be (0,4) and the range would be all reals. This would still be increasing over the domain, but with concave down.
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00:58:05 no idea.
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RESPONSE --> Huh?
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00:58:44 ** The domains and ranges of the function and its inverse would be switched. The inverse function should start out asymptotic to the negative x axis and for positive x should approach asymptote y = 4. Its concavity will therefore change at some point from concave up to concave down. The inverse function will lie between the vertical lines x = 0 and x = 4 and will be asymptotic to both.**
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RESPONSE --> Ah, I forgot the asymptotes.
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01:01:46 query text problem 1.3 #13 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts
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RESPONSE --> This is not 1.3.13. Still, here goes. H(30) = 10 means that at temperature t, that the height of the mercury in the thermometer is 10 (probably cm). Vertical intercept is where temperature is 0, while the horizontal intercept is where the height of the mercury is 0.
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01:01:54 the verticle ** vertical ** intercept is the temperature of the object when it is placed outside
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RESPONSE --> Ok.
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01:01:58 the horizontal intercept is the time when the object became the same temperature as the outside
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RESPONSE --> Ok.
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