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course Mth 173

͊xy~㛢assignment #007

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ލ܈y[HĒj

Calculus I

02-08-2007

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08:36:32

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 -->

The tangent line gives us a good approximation because it touches the curve at one point and is in the vicinity of the curve.

We would obtain the tangent line by getting the deriviative of y = kx^3, which is y' = 3kx^2. Then we can get two points of the curve and you will be able to determine the slope of the line by the rise/run.

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08:36:39

** 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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08:37:07

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 -->

dT/dt = k (T - 20)

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08:37:25

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 -->

I got the answer correct.

you got the correct equation; you should also have addressed, either in the problem or in your self-critique, the question about the graph.

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08:38:35

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 -->

0, 0

.5, .25

1, 1

1.5, 2.25

2, 4

0, 0

.25, .5

1, 1

2.25, 1.5

4, 2

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08:38:45

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 -->

I got the correct answer.

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08:39:41

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 -->

The points, when connected, cross the y=x line. It actually is a mirror imagine.

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08:39:49

** 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 -->

I got the answer correct.

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08:42:29

**** 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 -->

0, 0

1, 1

2, 4

3, 9

4, 16

5, 25

6, 36

7, 49

8, 64

9, 81

10, 100

11, 121

12, 144

If we reverse the table we would get the first column to be the square root and numbers ranging from 0 to 144 with the second column being the square root of and numbers ranging from 0 to 12.

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08:42:37

** 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 -->

I got the answer correct.

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08:43:36

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 every number is the square of another number.

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08:43:43

** 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 -->

I got the answer correct.

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08:44:16

What number would appear in the second column next to the number 4.31 in the first column?

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RESPONSE -->

4.31 ^ 2 = 18.5761

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08:44:24

** In the original table the second column would read 18.57, approx.. This is the square of 4.31. **

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RESPONSE -->

I got the answer correct.

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08:45:46

What number would appear in the second column next to the number `sqrt(18) in the first column?

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RESPONSE -->

The second column would be

sqrt of 18 = 4.242640687 ^ 2 = 18

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08:46:03

** 18 would appear in the second column because the square of sqrt(18) is 18. **

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RESPONSE -->

I got the answer correct.

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08:46:12

What number would appear in the second column next to the number `pi in the first column?

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RESPONSE -->

pi ^ 2

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08:46:21

** The number would be `pi^2 **

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RESPONSE -->

i got the answer correct.

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08:47:50

What would we obtain if we reversed the columns of this table?

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RESPONSE -->

The first column would be the numbers ^ 2 and the second column would be the numbers that the first column squared

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08:47:57

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 -->

I got the answer correct.

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08:48:51

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 -->

4.31 ^ 2 = 18.5761

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08:48:58

This number would be 4.31 squared,18.5761.

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RESPONSE -->

I got the answer correct.

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08:49:35

What number would appear in the second column next to the number

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RESPONSE -->

18.5761

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08:50:03

`pi^2 in the first column of this table?

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RESPONSE -->

The square root of pi

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08:50:12

STUDENT ANSWER: This number would be the square root, 'pi

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RESPONSE -->

I got the answer correct.

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08:51:19

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 -->

a negative number is not a number that can be squared.

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08:51:30

** 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 -->

I believe I got this answer correct.

You had the right idea. But a negative number can certainly be squared. However a negative number cannot be a square.

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08:54:10

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 2 (18) = x

2 ^ (4x) = 12

log 2 (12) = 4x

4096 = 4x

1024 = x

5 * 2 ^ x = 52

2 ^ x = 10.4

log 2 (10.4) = x

2 ^ (3x-4) = 9

log 2 (9) = 3x - 4

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08:55:58

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:

x = log{base 2}(10.4) so

x = 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 -->

I don't understand how the answer for 5 * 2 ^ x = 52

How does it become 2x? Would you not take 2 ^ x = 10.4 and then log 2 (10.4 ) = x?

That solution was incorrect, and your solution is correct. Good eye. Not many students point out that error.

I've corrected the solution in this document.

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08:57:47

14. Solve 2^(3x-5) + 4 = 0

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RESPONSE -->

2 ^ (3x-5) + 4 = 0

log 2 (0) + 4 = 3x - 5

log 2 (0) -1 = 3x

1 - 1 = 3x

0 = 3x

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08:57:58

**

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RESPONSE -->

There is not a solution here.

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08:58:47

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 -->

I need to work on the rules.

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08:58:49

**

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RESPONSE -->

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09:01:48

Solve 2^(1/x) - 3 = 0

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RESPONSE -->

2 ^ (1/x) - 3 = 0

2 ^ (1/x) = 3

log 2 ^ (1/x) = log 3

* 1/x * 1/x

log 2 / log 3 = x

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09:02:00

** 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 -->

I think I done this correctly.

You did. Good work.

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09:05:02

Solve 2^x * 2^(1/x) = 15

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RESPONSE -->

2 ^ x * 2 ^ 1/x = 15

2 ^ x + 1/x = 15

log 2 (x + 1/x) = log 15

x + 1/x = log 15/log 2

* 1x * 1x

x = log 15/log * 1x

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09:05:46

** 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 -->

I don't understand what I done wrong with the equation. What easy way is there to keep up with all of the rules?

You applied the rules correctly.

x + 1/x = log 15/log 2, as you say.

Your algebra after that point was, however, faulty. If you multiply x + 1 / x by x, you get x^2 + 1, not just x. (be sure also that you aren't confusing x + 1 / x with (x + 1) / x).

This gives you the equation x^2 + 1 = [log(15) / log(2) ] * x, which you have to recognize and solve as a quadratic equation.

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09:07:48

Solve (2^x)^4 = 5

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RESPONSE -->

2 ^ x ^ 4 = 5

log 2 (x ^ 4) = log 5

4 log 2 (x) = log 5

4 * x log 2 =log 5

4x log 2 = log 5

4x = log 5/log 2

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09:08:18

** 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 -->

I think that I went as far as I could with the equation. Can it be solved?

The last step is just to divide by 4 to get

x = log(5) / (4 log(2)).

This exact result can of course be approximated to any desired degree of precision by your calculator.

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09:11:42

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) is the cost for 10k square feet.

f ^ -1 (20 k) means that it costs f ^ -1 for 20k square feet

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09:12:17

** 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 -->

why is the second part of this question know money instead of square feet? what causes it to change?

The inverse function reverses the quantities. If f(x) represents cost f(x) vs. square feet x (thing of two columns, the first being square feet and the second cost), then f^-1(x) represents square feet f^-1(x) vs. cost x (the columns would reverse so that square feet appear in the first column and the cost in the second).

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09:14:22

problem 1.3.20. vert stretch y = x^2 by factor 2 then vert shift 1.

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RESPONSE -->

If we factor (multiply) by 2 then we get:

y = 2x ^ 2

Then if we shift on a graph vertically that means going up the y axis. It says to shift by 1 which means we go up the y axis by 1. Then the equation is

y = 2x ^ 2 +1

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09:14:32

** 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 -->

I got the answer correct.

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09:16:49

Give the equation of the function.Describe your sketch in detail.

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RESPONSE -->

y = 2x ^ 2 + 1

It shifts the curve up 1 point at every point of the curve.

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09:17:22

** 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 -->

I think I got this correct.

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09:23:05

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 -->

First find the value of g(x), which is 1

then apply 1 to f(x)

You don't substitute 1 into f(x), you substitute g(1). This is what you correctly describe below.

If you look at graph g(x) and look at the x axis at point 1 and figure there the line intercepts the sketch, it is 2 on the y axis.

Then we look on the f(x) graph and find 2 on the x axis and figure where the line intercepts and it is around .5 on the y axis.

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09:23:07

*&*& right problem? *&*&

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RESPONSE -->

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09:23:16

** 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 -->

I think I got it right.

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09:28:47

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 -->

I am confused on this problem

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09:29:47

** 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 -->

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

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09:30:21

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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09:30:23

no idea.

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RESPONSE -->

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09:30:26

** 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 -->

In a good self-critique you need identify the specific things you do and do not understand in the given solution, and either demonstrate your understanding or ask specific questions about what you don't understand. It doesn't accomplish the intended learning goals to simply say that you understand.

That way, once you have defined your difficulties I can help you address them.

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09:33:28

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 -->

The graph would intercept at the vertical when the item is first placed and the intercept at the horizontal is when the item gets the same temp as where it was placed.

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09:33:29

the verticle ** vertical ** intercept is the temperature of the object when it is placed outside

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RESPONSE -->

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09:33:32

the horizontal intercept is the time when the object became the same temperature as the outside

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RESPONSE -->

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09:33:47

*&*& H is the temperature, t is the clock time. H(30) is the temperature at clock time t = 30, so H(30) = 10 tells us that a clock time t = 30 the temperature was 10 degrees.

The vertical coordinate is the temperature, and the vertical intercept of the graph occurs when t = 0 so the vertical intercept gives us the temperature at clock time 0.

The vertical coordinate is the clock time, and the horizontal intercept occurs when H = 0, so the horizontal intercept gives us the clock time when temperature is 0. *&*&

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RESPONSE -->

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"

You did well overall up to the last few problems, which are fairly challenging conceptual problems. You should have self-critiqued those solutions in the indicated manner. You may still do so if you wish to copy the problems into a document and insert detailed self-critiques.