key stone problem… key stone problem… next set 20 © 2007 herbert i. gross
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Key Stone Problem…Key Stone Problem…
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Set 20© 2007 Herbert I. Gross
You will soon be assigned problems to test whether you have internalized the material
in Lesson 20 of our algebra course. The Keystone Illustration below is a
prototype of the problems you’ll be doing. Work out the problems 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.
Instructions for the Keystone Problem
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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© 2007 Herbert I. Gross
In program format the function f is defined by the following steps…
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Keystone Problem for Lesson 20Keystone Problem for Lesson 20
“f - program”Step 1 Start with xStep 2 Add 3Step 3 Multiply by 4Step 4 Add xStep 5 Multiply by 2Step 6 Subtract 20Step 7 Answer is f(x)
For what value of x is f(x) = 150?
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© 2007 Herbert I. Gross
When written in a step-by-step format the most direct way to find the inverse function
is to start with the last step (in this case it means to start with 150 in
Step 7) and then successively “undo”each of the preceding steps.
SolutionSolution
However, this process can face “road blocks”
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we see that this step is the first (and only) step in which the input is not the output of
the previous step. In other words, to “undo” Step 4 we would have to know the input of Step 1; but in the “undoing” process we do
not know what this number is.
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© 2007 Herbert I. Gross
Step 4 Add x
More specifically, if we look at
SolutionSolution
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© 2007 Herbert I. Gross
So our next approach is to replace the above sequence of steps by an equivalent set of steps for which each step can be “undone”. To this end we see that…
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SolutionSolution
“f - program”Step 1 Start with x xStep 2 Add 3 x + 3Step 3 Multiply by 4 4(x + 3)
Step 4 Add x4(x + 3) + x
Step 5 Multiply by 2 2(5x + 12)Step 6 Subtract 20 10x + 24 + -20
= 10x + 24
Step 1 Answer is f(x) 10x + 4
= 4x + 12 + x= 5x + 12
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© 2007 Herbert I. Gross
Thus, the function f can be represented more simply as…
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SolutionSolution
f(x) = 10x + 4
f(x) = 10x + 4
150 If we now replace f(x) by 150 in the equation we see that…
Subtracting 4 from both sides of the above equation we obtain…
146 = 10x
and if we now divide both sides of the above equation by 10 we see that…
14.6 = x
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Part of the solution requires that we check our answer.
To check that the answer we obtained in our equation is correct, we replace x by 14.6 in the “f-Program” and see that….
Notes
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© 2007 Herbert I. Gross
“f - program”Step 1 Start with x 14.6Step 2 Add 3 17.6Step 3 Multiply by 4 70.4
Step 5 Multiply by 2 170Step 6 Subtract 20 150Step 1 Answer is f(x) 150
Step 4 Add x 70.4 + 14.6 = 85
The fact that in its present form we cannot undo the above program doesn’t mean that the inverse of the program doesn’t exist. In particular in finding the solution to the
problem we showed that in algebraic format the “f-Program” was equivalent to…
Notes
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© 2007 Herbert I. Gross
f(x) = 10x + 4
In other words, the f-programNotes
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© 2007 Herbert I. Gross
“f - program” f(x) = 10x + 4Step 1 Start with x Start with xStep 2 Add 3 Multiply by 10Step 3 Multiply by 4 Add 4
Step 5 Multiply by 2Step 6 Subtract 20Step 7 Answer is f(x)
Answer is f(x)Step 4 Add x
…can be paraphrased into the simpler f(x) = 10x + 4 program…
And in this form we see that the “undoing” program is…
Notes
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© 2007 Herbert I. Gross
Step 1 Start with xStep 2 Multiply by 10Step 3 Add 4
The “undoing” ProgramThe “undoing” ProgramStep 4 Answer is f(x)
Answer is x Step 4Divide by 10 Step 3Subtract 4 Step 2
Start with f(x) Step 1
The ProgramThe Program
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© 2007 Herbert I. Gross
The “undoing” program is the inverse function of f.
In other words, f-1 is defined by…
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“f-1 - program”Step 1 Start with x xStep 2 Subtract 4 x – 4
Step 4 Answer is f-1(x) f-1(x) = 1/10 (x – 4)
Step 3 Divide by 10(x – 4) ÷ 10
= 1/10 (x – 4)
Notes
With respect to the given problem, we wanted to know the value of x if f(x) = 150. The answer is given by x = f-1(150); and from Step 4 above we see that…
Notes
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© 2007 Herbert I. Gross
Step 4 Answer is f-1(x) f-1(x) = 1/10(x – 4)
f-1( x ) = 1/10( x – 4 )150
= 1/10( 146 )
= 14.6
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150
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© 2007 Herbert I. Gross
Remember that by definition f-1(f(x)) = x for each input x. As a check in the present
example, we see that…
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““f - program”f - program”Step 1 Start with x xStep 2 Add 3 x + 3Step 3 Multiply by 4 4(x + 3)
Step 5 Multiply by 2 2(5x – 12)Step 6 Subtract 20 10x + 24 + -20
= 10x + 24
Step 7 Answer is f(x) 10x + 4
= 5x + 12
Notes
Step 4 Add x 4x + 12 + x
Step 8 Subtract 4 10x + 4 – 4 = 10xStep 9 Divide by 10 10x ÷ 10
Step 10 Write the answer x
““ff-1-1 - program” - program”
xx
xx
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© 2007 Herbert I. Gross
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Starting with the function we
called f, we draw its graph, namely the line L whose
equation is y = f(x) = 10x + 4.
(0,4)
(1,14)
(14,1)(4,0)
(-1,-6)
(-6,-1)
y = x
y = f(x) = 10x + 4
y = f-1(x) = 1/10x – 4
We then draw the line y = x,
The resulting line represents the
function f-1 where f-1(x) = 1/10(x – 4).
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and finally, we reflect L about the line
y = x.
next LGeometric SummaryGeometric Summary
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© 2007 Herbert I. Gross
There are times when even with more advanced knowledge of algebra we cannot
explicitly find f-1. In such cases one can always resort to trial and error.
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Trial and Error
For example, to find the value of x for which f(x) = 150 we could have just tried a few
values of x to see what was ”going on”.
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© 2007 Herbert I. Gross
For example, suppose we used as our first guess that x = 10. We would see that…
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““f - program”f - program”Step 1 Start with x 10Step 2 Add 3 13Step 3 Multiply by 4 52
Step 5 Multiply by 2 124Step 6 Subtract 20 120Step 7 Answer is f(x) 120 (which is < 150)
Step 4 Add x 62
Trial and Error
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© 2007 Herbert I. Gross
We might then try a greater value for x, say x = 20. In this case we would see that…
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““f - program”f - program”Step 1 Start with x 20Step 2 Add 3 23Step 3 Multiply by 4 92
Step 5 Multiply by 2 224Step 6 Subtract 20 204Step 7 Answer is f(x) 204 (which is > 150)
Step 4 Add x 112
Trial and Error
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© 2007 Herbert I. Gross
Thus, we may conclude that since f(10) is less than 150 and f(20) is greater than 150,
there has to be at least one value of x between 10 and 20 for which f(x) = 150
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Trial and Error
If the above estimate was not close enough to suit our purpose, we could pick a value for
x that is, say, halfway between 10 and 20 (that is 15).
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© 2007 Herbert I. Gross
If we let x = 15 we see that…
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““f - program”f - program”Step 1 Start with x 15Step 2 Add 3 18Step 3 Multiply by 4 72
Step 5 Multiply by 2 174Step 6 Subtract 20 154Step 7 Answer is f(x) 154
Step 4 Add x 87
Trial and Error
The fact that 154 is “just a little bigger” than 150 tells us that the correct value of x must be a
“little less” than 15.
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© 2007 Herbert I. Gross
To be on the safe side we might next try x = 14 as
our guess. In this case we would obtain…
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““f - program”f - program”Step 1 Start with x 14Step 2 Add 3 17Step 3 Multiply by 4 68
Step 5 Multiply by 2 164Step 6 Subtract 20 144Step 7 Answer is f(x) 144
Step 4 Add x 82
Trial and Error
…and 144 is “just a little less” than 150.
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© 2007 Herbert I. Gross
We could then conclude that since f(14) = 144 and f(15) = 154 that
there is a value of x that is between 14 and 15 for which f(x) = 150.
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Trial and Error
Although the process might seem tedious, we could keep repeating it until we obtained a value for x that was sufficiently accurate
for our purpose.
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© 2007 Herbert I. Gross
For example, if we now were to let x = 14.5, we would see that…
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““f - program”f - program”Step 1 Start with x 14.5Step 2 Add 3 17.5Step 3 Multiply by 4 70
Step 5 Multiply by 2 169Step 6 Subtract 20 149Step 7 Answer is f(x) 149
Step 4 Add x 84.5
Trial and Error
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© 2007 Herbert I. Gross
Trial and Error
Since 149 is close to 150, we would know that the desired value of x is between 14.5 and 15, but probably much closer to 14.5.
We could continue with this “refining” process if this estimate was not sufficiently
accurate for our purposes.
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© 2007 Herbert I. Gross
Final Note
While the trial-and-error method allows us to approximate the answer to as great a
degree of accuracy as we wish, it does not address the issue of whether there is more
than one answer to the problem.
Thus, one advantage of using algebra whenever possible is that it tells us more
about the uniqueness of the answer.
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