# Key Stone Problem… Key Stone Problem… next Set 20 © 2007 Herbert I. Gross.

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Key Stone Problem… Key Stone Problem… next 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 next © 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. next © 2007 Herbert I. Gross

next © 2007 Herbert I. Gross In program format the function f is defined by the following steps… next Keystone Problem for Lesson 20 “f - program” Step 1Start with xStep 2Add 3Step 3Multiply by 4Step 4Add xStep 5Multiply by 2Step 6Subtract 20Step 7Answer is f(x) For what value of x is f(x) = 150? next

© 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. Solution However, this process can face “road blocks” next

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. © 2007 Herbert I. Gross Step 4Add x More specifically, if we look at Solution

next © 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… next Solution “f - program”Step 1Start with xxStep 2Add 3x + 3Step 3Multiply by 44(x + 3) Step 4Add x 4(x + 3) + x Step 5Multiply by 22(5x + 12)Step 6Subtract 2010x + 24 + - 20 = 10x + 24 Step 1Answer is f(x)10x + 4 = 4x + 12 + x = 5x + 12

next © 2007 Herbert I. Gross Thus, the function f can be represented more simply as… next Solution 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 next

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 next © 2007 Herbert I. Gross “f - program”Step 1Start with x14.6Step 2Add 317.6Step 3Multiply by 470.4Step 5Multiply by 2170Step 6Subtract 20150Step 1Answer is f(x)150Step 4Add x70.4 + 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 next © 2007 Herbert I. Gross f(x) = 10x + 4

In other words, the f-program Notes next © 2007 Herbert I. Gross “f - program”f(x) = 10x + 4 Step 1Start with x Step 2Add 3 Multiply by 10 Step 3Multiply by 4 Add 4 Step 5Multiply by 2Step 6Subtract 20Step 7Answer is f(x) Step 4Add x …can be paraphrased into the simpler f(x) = 10x + 4 program…

And in this form we see that the “undoing” program is… Notes next © 2007 Herbert I. Gross Step 1Start with xStep 2Multiply by 10Step 3Add 4 The “undoing” Program Step 4Answer is f(x)Answer is xStep 4Divide by 10Step 3Subtract 4Step 2Start with f(x)Step 1 The Program next

© 2007 Herbert I. Gross The “undoing” program is the inverse function of f. In other words, f -1 is defined by… next “f -1 - program”Step 1Start with xxStep 2Subtract 4x – 4Step 4Answer is f -1 (x)f -1 (x) = 1/10 (x – 4) Step 3Divide 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… next © 2007 Herbert I. Gross Step 4Answer is f -1 (x)f -1 (x) = 1 / 10 (x – 4) 150 = 1 / 10 ( 146 ) = 14.6 next 150

next © 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… next “f - program” Step 1Start with xxStep 2Add 3x + 3Step 3Multiply by 44(x + 3)Step 5Multiply by 22(5x – 12)Step 6Subtract 2010x + 24 + - 20 = 10x + 24 Step 7Answer is f(x)10x + 4 = 5x + 12 Notes Step 4Add x4x + 12 + xStep 8Subtract 410x + 4 – 4 = 10xStep 9Divide by 1010x ÷ 10Step 10Write the answerx “f -1 - program” x x

next © 2007 Herbert I. Gross next 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 / 10 x – 4 We then draw the line y = x, The resulting line represents the function f -1 where f -1 (x) = 1 / 10 (x – 4). next and finally, we reflect L about the line y = x. L Geometric Summary

next © 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. next 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”.

next © 2007 Herbert I. Gross For example, suppose we used as our first guess that x = 10. We would see that… next “f - program” Step 1Start with x10Step 2Add 313Step 3Multiply by 452Step 5Multiply by 2124Step 6Subtract 20120Step 7Answer is f(x)120 (which is < 150)Step 4Add x62 Trial and Error

next © 2007 Herbert I. Gross We might then try a greater value for x, say x = 20. In this case we would see that… next “f - program” Step 1Start with x20Step 2Add 323Step 3Multiply by 492Step 5Multiply by 2224Step 6Subtract 20204Step 7Answer is f(x)204 (which is > 150)Step 4Add x112 Trial and Error

next © 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 next 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).

next © 2007 Herbert I. Gross If we let x = 15 we see that… next “f - program” Step 1Start with x15Step 2Add 318Step 3Multiply by 472Step 5Multiply by 2174Step 6Subtract 20154Step 7Answer is f(x)154Step 4Add x87 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. next

© 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… next “f - program” Step 1Start with x14Step 2Add 317Step 3Multiply by 468Step 5Multiply by 2164Step 6Subtract 20144Step 7Answer is f(x)144Step 4Add x82 Trial and Error …and 144 is “just a little less” than 150. next

© 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. next 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.

next © 2007 Herbert I. Gross For example, if we now were to let x = 14.5, we would see that… next “f - program” Step 1Start with x14.5Step 2Add 317.5Step 3Multiply by 470Step 5Multiply by 2169Step 6Subtract 20149Step 7Answer is f(x)149Step 4Add x84.5 Trial and Error

next © 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.

next © 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. next