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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
© 2007 Herbert I. Gross
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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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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Step 1 The input is xStep 2Step 3Step 4Step 5Step 6Step 7
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
© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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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)
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
© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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…
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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
Step 1 The input is xStep 2Step 3Step 4Step 5Step 6Step 7
Add 3Multiply by 2
Add x (14)Divide by 3Subtract 1
The output is f(x)
141734481615 1515
© 2007 Herbert I. Gross
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Step 1 The input is xStep 2Step 3Step 4Step 5Step 6Step 7
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
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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…
© 2007 Herbert I. Gross
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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…
© 2007 Herbert I. Gross
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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.
© 2007 Herbert I. Gross
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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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© 2007 Herbert I. Gross
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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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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© 2007 Herbert I. Gross
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Just by using our knowledge of arithmetic, we may compute f(x) for
various values of x.
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) 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.
SummarySummarynext
© 2007 Herbert I. Gross
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
© 2007 Herbert I. Gross