Key Stone Problem… Key Stone Problem… Set 17 Part 2 © 2007 Herbert I. Gross next

Key Stone Problem…

Set 17 Part 2

© 2007 Herbert I. Gross

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You will soon be assigned problems to test whether you have internalized the material in Lesson 17 Part 2 of our algebra course.

The Keystone Illustration below is a prototype of the problems you’ll be doing.

Work out the problem on your own. Afterwards, study the detailed solutions we’ve provided. In particular, notice that several different ways are presented that could be used to solve each problem.

Instruction for the Keystone Problem


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As a teacher/trainer, it is important for you to understand and be able to respond

in different ways to the different ways individual students learn. The more ways

you are ready to explain a problem, the better the chances are that the students

will come to understand.© 2007 Herbert I. Gross

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Keystone Problem for Keystone Problem for Lesson 17 Part 2Lesson 17 Part 2

Written in the “Program” format, the function f is defined by…

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What is the value of x if f(x) = 15?

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Step 1 The input is xStep 2Step 3Step 4Step 5Step 6Step 7

Add 3Multiply by 2

Add xDivide by 3Subtract 1

The output is f(x)

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Solution for the ProblemSolution for the ProblemPerhaps the most natural approach would be to start with Step 7, and then undo each step until we arrive back at Step 1. Things work well until we get to step 4 (i.e. Add x).

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That is, we first undo Step 6 by adding 1 to 15 to obtain 16.

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Next we undo Step 5 by multiplying 16 by 3 to obtain 48. However, when we try to undo Step 4 by “unadding x”; we are stymied by the fact that we don't know what the value of x is!

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Solution for the ProblemSolution for the Problem

So the next approach might be to try to paraphrase the program into an

equivalent program in which each step is able to be undone. And to keep the algebraic expressions as simple as

possible, we will use the rules of algebra to simplify each step before we proceed

to the next step.

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To this end we see that…

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Since f(x) = x + 1, the fact that f(x) = 15 means that x + 1 = 15, and hence…

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Add 3Multiply by 2

Add xDivide by 3

Subtract 1 (Add -1)The output is f(x)

xx + 32(x + 3)(2x + 6) + x(3x + 6) ÷ 3(x + 2) + -1 f(x) = x + 1

= 2x + 6= (2x + x) + 3= 1/3 (3x + 6) = x + 2

= x + (2 + -1)

x =x = 1414

nextnextnextnextnextnext Solution for the ProblemSolution for the Problem

= 3x + 6

= x + 1


Paraphrasing

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This exercise emphasizes the importance of paraphrasing. More specifically, there is a tendency to identify the inverse function

with the concept of step-by-step “undoing”,

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However, in this case the step-by-step undoing process breaks down when we try

to undo Step 4 (Add x).

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If we were to convert the given program into the f(x) format without simplifying each step before proceeding to the next step, the result would have been…

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Add 3Multiply by 2


The output is f(x)

xx + 32(x + 3)2(x + 3) + x[2(x + 3) + x] ÷ 3[2(x + 3) + x] ÷ 3 – 1 f(x) = [2(x + 3) + x] ÷ 3 – 1

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Note


Notenext

So what we have shown is that the expression [2(x + 3) + x] ÷ 3 – 1is equivalent to the much simpler expression x + 1. In terms of an equation…

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[2(x + 3) + x] ÷ 3 – 1 = x + 1[2(x + 3) + x] ÷ 3 – 1 = x + 1

…is a true statement for every value of x.

Recall that an equation that is satisfied by Recall that an equation that is satisfied by every value of x is called an every value of x is called an identityidentity..

Identities will be studied in more detail in Lesson 18.

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In terms of the program model, it means that we may replace the given program…

Step 1 The input is xStep 2Step 3

Add 1The output is f(x)

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…by the much simpler, but equivalent, program…


Add 3Multiply by 2


The output is f(x)

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Note

In this form, it is easy to see that we “undo” the output simply by subtracting 1. Hence, in order for the output to be 15, the

input has to be 14.

Step 1 The input is xStep 2Step 3

Add 1The output is f(x)

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We can check that our answer

is correct by replacing x by 14

in the original program and

validating that in this case the output is 15.

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Checking our SolutionChecking our Solution


Add 3Multiply by 2

Add x (14)Divide by 3Subtract 1

The output is f(x)

141734481615 1515


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Add 3Multiply by 2

Add x (14)Divide by 3Subtract 1

The output is f(x)

141734481615 1515

In other words while we can't reverse the steps in the program by the undoing method, we can undo the entire program with the single step “Subtract 1”.

Step 8 Subtract 1 1414

1414

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Trial and Error

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This idea is particularly important when the algebra might be too complicated for us to

paraphrase.

The fact that algebra is unarithmetic means that we can use arithmetic and trial and

error to find the input if we know the program (function) and the output.

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Notenext For example, suppose…next

f(x) = x5 + x

…and we want to know the value of x for which f(x) = 30. That is: we would like to find a value of x for which x5 + x = 30.

If we replace x by 1, we see that…

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x 5 + x =1 21

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And now if we replace x by 2, we see that…

x 5 + x =2 342

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Notenext Since 2 is less than 30, but 34 is greater than 30, there must be a value of x between 2 and 34 for which x5 + x = 30.Moreover, since the difference between 30 and 34 is much less than the difference between 30 and 2, we suspect that the desired value of x is closer in value to 2 than to 1. So as a hunch, we might replace x by 1.9. (Since computing x5 longhand is quite tedious, use a calculator with an xY key.)

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nextnextIn which case we would obtain…

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x 5 + x =1.91.9

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9 8 7 +6 5 4 -3 2 1 ×%

xy =÷

On/off √0 .

xy

1

=

9

.

1.9 524.76099In words: enter “1.9”;press the

xy key; enter“5”;

press the = key

Then enter the plus key

and 1.9.

Press the = key.

5+

1

.

9

=

1.926.6609926.66099

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= 28.19506…

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x 5 + x =

+ 1.95 1.955

Since 26.66099 is still less than 30, we know that the desired value of x must be greater than 1.9, but still less than 2. So our next hunch might be to replace x by 1.95 in which case we would obtain…

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30.14506…

1.95 1.95

30.14506…


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= 27.47948…

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x 5 + x =

+ 1.94 1.945

Since x = 1.95 gives us an output which is slightly greater than 30, our next guess would be a value of x which is slightly less than 1.95. So suppose we choose x = 1.94, we would obtain…

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29.41948…

1.94 1.94

29.41948…


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We see that x = 1.94 gives an output that is less than 30; and since x = 1.95 gave us an

output greater than 30, we know that the desired value of x is greater than 1.94 but

less than 1.95.

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We now have a fairly good estimate of the value of x for which x5 + x = 30. However, if necessary, we could continue this process

to obtain an even better estimate.


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x x5 x5 + x (=30?) Results

2 32 34 Too Big1.95 28.19506… 30.14506… Too Big

1.949 28.12284… 30.07184… Too Big1.9484 28.07957… 30.02797… Too Big

1 1 2 Too Small1.9 24.76099 26.66099 Too Small1.94 27.47948… 29.41948… Too Small

1.948 28.05076… 29.99876… Too Small---------- ----------- 30 Perfect!

Thus, the chart shows us that the desired value of x is between 1.948 and 1.9484. Hence, we

may say that rounded off to the nearest thousandth, the exact value of x is 1.948.

nextnextnextnextnextnextnextnext Charting Our Estimates

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As an application to our Keystone Problem, suppose we didn’t know algebra, but were confronted with the problem…

With f defined as shown, what is the value of x if f(x) = 15?

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Add 3Multiply by 2


The output is f(x)

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Just by using our knowledge of arithmetic, we may compute f(x) for

various values of x.


Add 3Multiply by 2


The output is f(x) 15

10132636121111

20234666222121

12153042141313

14173448161515

16193854181717

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Since an input of 10 gave us an output that was less than 15 and an input of 20 gave us an output that was greater than 15, we knew that the correct input had to be between 10 and 20. We then obtained a similar result when we used 12 and 16

as the inputs, thus telling us that the correct input had to be between 12 and

16; and eventually we found that the correct input was 14.

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However, when we use trial and error to find an answer, we cannot be sure whether

or not there are other answers. On the other hand, using algebra in this example we were able to show that x = 14 was the

only correct answer.

Moreover, when we know how to apply algebra, it is usually far less tedious than resorting to trial-and-error techniques.

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SummarySummary


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Key Stone Problem… Key Stone Problem… Set 17 Part 2 © 2007 Herbert I. Gross next