query 7

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course Mth 173

6/9 12:40 am

007. `query 7

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Question: `q 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.3

** If we know that y=kx^3, as in the sandpile model, we can find the derivative as y = 3kx^2.

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Your solution:

If you take a point on the graph and substitute the y and x values you can find the k or slope of the value. After finding the slope you find the derivative function which gives the x or y value you’re missing for a point on the tangent line

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Given Solution:

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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Question: `q 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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Your solution:

dT/dt = k (T-20)

a straight line increasing. Because the values of temperature increase

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Given Solution:

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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Question: `q 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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Your solution:

X I x^2

0 0

.5 .25

1 1

1.5 2.25

2 4

Inverse table:

0 0

.25 .5

1 1

2.25 1.5

4 2

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Given Solution:

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).

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?

** 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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Question: `q **** 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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Your solution:

0 0

1 1

4 2

9 3

16 4

25 5

36 6

49 7

64 8

81 9

100 10

121 11

144 12

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Given Solution:

** 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. **

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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Question: `q ** 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.

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Your solution:

All possible x values squared could have an inverse

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Given Solution:

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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Question: `q What number would appear in the second column next to the number 4.31 in the first column?

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Your solution:

If you square the 4.31 you would get 18.58

If you take the square root of 4.31 you get 2.08

confidence rating #$&*:3

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Given Solution:

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

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

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Your solution:

4.24

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Given Solution:

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

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This makes sense

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

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Your solution:

9.87

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Given Solution:

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

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Question: `q What would we obtain if we reversed the columns of this table?

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Your solution:

We would get the inverse. The squares would be in the x column now and the square roots in the y column

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Given Solution:

STUDENT ANSWER: We would obtain the inverse, the square roots of the squares being in the y colum and the squared numbers being in the x column.

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Question: `q 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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Your solution:

18.58

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Given Solution:

This number would be 4.31 squared,18.5761.

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Question: `q What number would appear in the second column next to the number `pi^2 in the first column of this table?

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Your solution:

The square root of pi ^2 or pi

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Given Solution:

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

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Question: `q What number would appear in the second column next to the number -3 in the first column of this table?

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Your solution:

The square root can not be taken of a negative number

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Given Solution:

There is no such number. No real number has a square equal to -3, since the square of any number which is positive or negative is the product of two numbers of like sign and is therefore positive.

Put another way: sqrt(-3) is not a real number, since the square of a real number cannot be negative. **

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Question: 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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Your solution:

Log (2^x) = log 18

X log 2 = 1.26

X= 4.17

2^ (4x) = 12

Log (2^(4x)) = log 12

4x log 2 = log 12

4x= 3.58

X= .90

5* 2^x = 52

log 5 * x log 2= log 52

x log 2 = 2.455

x= 8.16

2^ (3x-4) = 9

2^3x + 2^-4 = 9

3x log 2 + -4 log 2 = 9

3x log 2 - 1.2 = 9

3x log 2 = 10.2

3x= 33.88

X= 11.29

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Given Solution:

b^x = c is translated into logarithmic notation as log{base b}(c) = x. So:

2^x = 18 translates directly to log{base 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)

You can easily evaluate this and the preceding solution on your calculator.

2^(3x-4) = 9 translates to log{base 2}(9) = 3x - 4.

Solving for x we get

x = (log(base 2)(9) + 4) / 3.

This can be evaluated using a calculator.

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Question: `q 14. Solve 2^(3x-5) + 4 = 0

**

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Your solution:

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

3x log 2 + -5log 2 + log4 = 0

3x log 2 -1.51 + .602 = 0

3x log 2 = .908

3x= 3.02

X= 1.005

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Given Solution:

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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Question: `q Solve 2^(1/x) - 3 = 0

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Your solution:

(1/x) log2 = log 3

1/ x= log 3/ log 2

X = log 2/ log3

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Given Solution:

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

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Your solution:

X log 2 * (1/x) log 2 = log 15

X * (1/x) log 2 = log 15/ log 2

X * (1/x) = log15/ log 2/ log2

X (1/x) = 12.98

X^2 + 1 = 12.98x

X^2 -12.98x +1 = 0

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Given Solution:

** 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. **

Solve (2^x)^4 = 5

** 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.**

STUDENT QUESTION

Even with the solution I am missing a step as I work from x+(1/x)=log15/log2 to the final solution?

INSTRUCTOR RESPONSE

As stated, you multiply both sides of that equation by x.

x * ( x + 1/x) = x^2 + 1, and multiplying the other side by x gives the obvious expression, so you get

x^2 + 1 = [log(15) / log(2) ] * x

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Question: `q problem 1.3.20 5th; 1.3.22 4th. C=f(A) = cost for A sq ft. What do f(10k) and f^-1(20k) represent?

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Your solution:

C = f(A)

F(10k) you can use to find the cost when the area if 10,000 square feet

The inverse you can use to find square foot when the cost is 20,000

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Given Solution:

** 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. **

STUDENT COMMENT

Still not positive about the - 1.

INSTRUCTOR RESPONSE

f ^-1 (x) is the notation for the inverse function.

If x is quantity A and f(x) is the value of quantity B, then when you invert the function x becomes quantity B and f ^-1 (x) becomes quantity A.

In the original function x is the area and f(x) is the cost. When inverted to the form f ^-1 (x), x becomes the cost and f ^-1 (x) the area.

You can think of inverting a function in terms of switching the columns of a table.

We can also think of inverting a function in terms of switching the x and y coordinates on a graph, which reflects the graph through the line y = x.

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Question: `q problem 1.3.38 4th edition (problem omitted from 5th edition but everyone should do this problem). Write an equation for the function if we vertically stretch y = x^2 by factor 2 then vertically shift the graph 1 unit upward.

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Your solution:

Y= 2(x^2) +1

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Given Solution:

** 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. **

STUDENT QUESTION

I graphed the solution for vertically stretching and I see the change in the graph but I am still confused on the idea of

Vertically stretching.

INSTRUCTOR RESPONSE

This is summarized at the very beginning of the section in the text.

When you multiply a function by c, you move every point | c | times as far from the x axis. It's as if you grabbed along of the top and bottom of the graph and stretched it out factor c (if | c | < 1 it's actually as if you compressed the graph). If c is negative the graph also reflects through the x axis.

This is a typical precalculus topic. If your precalculus or analysis course didn't cover this you might want to consider at least reading through, and perhaps working through at least some of the relevant parts of the following documents, which among other things provide a detailed introduction to understanding the concepts of stretching and shifting graphs:

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc1/pc1_qa5.htm

http://vhcc2.vhcc.edu/pc1fall9/Assignments/assignment_98126_function%20_families.htm

http://vhcc2.vhcc.edu/pc1fall9/pc1/basic_point_graphs_identifying_equation.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc1/pc1_qa6.htm

http://vhcc2.vhcc.edu/pc1fall9/basic_function_families/basic_function_families.htm<

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Question: `q Give the equation of the function. Describe your sketch in detail. Explain what effect, if any, it would have on the graph if we were to reverse the order of the transformations.

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Your solution:

Y = 2(x^2) + 1

The graph was stretched by a factor of 2 then up by 1

The result is a parabola which is concave up with vertex at point (0,1). After the shifts the points become, (-1, 3) and (1, 3).

If reversed, then the graph is shifted upward 1 unit before being stretched vertically by 2. When stretched, the points become (0, 2). The points (-1, 1) and (1, 1) shift to (-l, 2) and (1, 2), and the stretch then moves them to (-1, 4) and (1, 4).

The function becomes y = 2 ( x^2 + 1) =

2 x^2 + 2.

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Given Solution:

** The function would be y = f(x) = 2 x^2 + 1. The factor 2 stretches the y = x^2 parabola vertically 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, so that the x = +-1 points lie 2 units higher than the vertex, at the points (-1, 3) and (1, 3).

If the transformations are reversed then the graph is shifted upward 1 unit before being stretched vertically by factor 2. The vertex, for example, shifts to (0, 1) then when stretched moves to (0, 2). The points (-1, 1) and (1, 1) shift to (-l, 2) and (1, 2), and the stretch then moves them to (-1, 4) and (1, 4).

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. The vertex for the reversed pair of transformations lies 1 unit higher than the vertex for the original pair. The shapes of both graphs are parabolic, and are identical except. The reversed pair of transformations results in a parabola that is everywhere 1 unit 'higher' than that associated with the first pair. **

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Question: `q problem 1.3.45 5th; 1.3.43 4th (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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Your solution:

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Given Solution:

** 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 between 0 and 1, closer to 0. The value is about .3, give or take .1.. **

We can summarize this using function notation:

• f(g(1)) = f(2) = .3 (estimated)

Further note: It appears from the graph that g(1) might be a little greater than 2, in which case f(g(1)) might be closer to .4.

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@&

&#You need a detailed self-critique here.

&#

*@

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Question: `q problem Graph the function f(x) = x^2 + 3^x for x > 0.

Decide if this function has an inverse. If so, find the approximate value of the inverse function at x = 20.

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Your solution:

I don’t think this function would have an inverse because the graph would be a parabola

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Given Solution:

** 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 + 3^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. **

BEGINNING OF STUDENT SOLUTION

I started by graphing the function of y=x^2+3^x and y=x.

INSTRUCTOR RESPONSE

You don't say how you constructed the graph. If you used a graphing calculator, your solution would not be acceptable. It would be OK to use the calculator to check out your construction, but you would have to explain your construction of the graph.

One description of the construction:

From knowledge of basic power and exponential functions, you should know that x^2 and 3^x are both increasing functions for x > 0 (x^2 is decreasing for x < 0, but for the purpose of this question that doesn't matter).

It should be clear why this is so. The greater the positive number you square, the greater the result. And the greater the value of x, the greater the power to which you raise 3, so the greater will be the value of 3^x.

The graph of x^2 goes through (0, 0) and (1, 1). The graph of 3^x goes through (0, 1) and (1, 3). As x values continue to increase, the value of x^2 quickly becomes insignificant compared to that of 3^x (e.g., for x = 4 the function x^2 takes value 16 while 3^x takes value 81; for x = 8 we have x^2 = 64 and 3^x = 8261 (check my mental arithmetic on that one) ).

So the graph of x^2 + 3^x passes through (0, 1) and (1, 4), and continues increasing more and more quickly as x increases.

The resulting function is clearly invertible.

It would also be OK to make a table. You should be able to do enough mental arithmetic to do so without relying on a calculator. Certainly you can square the integers from, say, 0 to 5. And you should be able to start with 3 and triple your result, repeating at least 5 times.

x x^2 3^x x^2 + 3^x

0 0 1 1

1 1 3 4 4

2 4 9 13 13

3 9 27 36 36

4 16 81 77 77

5 25 343 368 368

A small portion of the graph is depicted below:

The values of x and x^2 + 3^x can easily be reversed, giving us a partial table of the inverse function:

x inv fn

1 0

4 1

13 2

37 3

77 4

368 5

It should be clear that the inverse function is single-valued (i.e., there is are no two x values greater than 0 for which the inverse function takes the same value).

A partial graph of the inverse function (not to exactly the same scale as the original) is depicted below:

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Self-critique (if necessary):

I graphed the function on a calculator and thought the function resulted in a parabola, after reading your explanation I realized that the graph does have an inverse and that it is very easy to find

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Self-critique Rating:2

@&

You should never rely on the calculator to construct a graph. Use it to check your construction, but it is not appropriate to use the calculator to replace analysis.

If x > 0, both x^2 and 3^x are strictly increasing, and doing so at an increasing rate. Their sum therefore increases at an increasing rate. From this is it clear that the given function does have an inverse when its domain is restricted to x > 0.

*@

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Question: `q If H = f(t) describes the temperature H of an object at clock time t, then what does it mean to say that H(30)=10? What information would you get from thevertical and horizontal intercepts of the graph of the function?

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Your solution:

If H = f(t) then h is a function of time

Temperature is dependent upon the independent variable of time

H(30) = 10 means at time 30 the temperature was 10 degrees

You would get the (temperature, time)

So at 10 degrees you would get the horizontal value of time

And at time 30 you would get the vertical value of temperature

confidence rating #$&*:2

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Given Solution:

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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Self-critique (if necessary):

OK

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Self-critique Rating:OK

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good work. See my notes and let me know if you have questions. &#