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

2. 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 next

3. 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 next

4. © 2007 Herbert I. Gross Keystone Problem for Lesson 17 Part 2 Written in the “Program” format, the function f is defined by… next What is the value of x if f(x) = 15? Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x Divide by 3 Subtract 1 The output is f(x)

5. next © 2007 Herbert I. Gross Solution for the Problem Perhaps 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). next  That is, we first undo Step 6 by adding 1 to 15 to obtain 16. next  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! next

6. © 2007 Herbert I. Gross Solution 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.

7. next © 2007 Herbert I. Gross To this end we see that… next Since f(x) = x + 1, the fact that f(x) = 15 means that x + 1 = 15, and hence… Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x Divide by 3 Subtract 1 (Add - 1) The output is f(x) x x + 3 2(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 = 14 next Solution for the Problem = 3x + 6 = x + 1

8. © 2007 Herbert I. Gross Paraphrasing next 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”, However, in this case the step-by-step undoing process breaks down when we try to undo Step 4 (Add x).

9. © 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… next Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x Divide by 3 Subtract 1 The output is f(x) x x + 3 2(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 next Note

10. © 2007 Herbert I. Gross Note next So what we have shown is that the expression [2(x + 3) + x] ÷ 3 – 1 is equivalent to the much simpler expression x + 1. In terms of an equation… [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 every value of x is called an identity. Identities will be studied in more detail in Lesson 18. next

11. © 2007 Herbert I. Gross In terms of the program model, it means that we may replace the given program… Step 1The input is x Step 2 Step 3 Add 1 The output is f(x) next …by the much simpler, but equivalent, program… Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x Divide by 3 Subtract 1 The output is f(x)

12. next © 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 1The input is x Step 2 Step 3 Add 1 The output is f(x)

13. next © 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. next Checking our Solution Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x (14) Divide by 3 Subtract 1 The output is f(x) 14 17 34 48 16 15 15

14. © 2007 Herbert I. Gross next Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x (14) Divide by 3 Subtract 1 The output is f(x) 14 17 34 48 16 15 15 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 8Subtract 114 14 next

15. © 2007 Herbert I. Gross Trial and Error next 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.

16. © 2007 Herbert I. Gross Note next For example, suppose… f(x) = x 5 + 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 x 5 + x = 30. If we replace x by 1, we see that… next x 5 + x = 1 2 1 next And now if we replace x by 2, we see that… x 5 + x = 2 34 2 next

17. © 2007 Herbert I. Gross Note next 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 x 5 + 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 x 5 longhand is quite tedious, use a calculator with an x Y key.)

18. © 2007 Herbert I. Gross next In which case we would obtain… x 5 + x =1.9 next 987+ 654- 321× % xyxy = ÷ On/off √ 0. xyxy 1 = 9. 1.9 5 24.76099 In words: enter “1.9”; press the x y key; enter“5”; press the = key Then enter the plus key and 1.9. Press the = key. 5 + 1. 9 = 1.926.66099 next

19. © 2007 Herbert I. Gross next = 28.19506… x 5 + x = + 1.95 1.95 5 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… next 30.14506… 1.95 30.14506…

20. © 2007 Herbert I. Gross next = 27.47948… x 5 + x = + 1.94 1.94 5 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… next 29.41948… 1.94 29.41948…

21. © 2007 Herbert I. Gross next 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. We now have a fairly good estimate of the value of x for which x 5 + x = 30. However, if necessary, we could continue this process to obtain an even better estimate.

22. © 2007 Herbert I. Gross next xx5x5 x 5 + x (=30?)Results 23234Too Big 1.9528.19506…30.14506…Too Big 1.94928.12284…30.07184…Too Big 1.948428.07957…30.02797…Too Big 112Too Small 1.924.7609926.66099Too Small 1.9427.47948…29.41948…Too Small 1.94828.05076…29.99876…Too Small ---------------------30Perfect! 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. next Charting Our Estimates

23. next © 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? next Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x Divide by 3 Subtract 1 The output is f(x)

24. next © 2007 Herbert I. Gross next Just by using our knowledge of arithmetic, we may compute f(x) for various values of x. Step 1The input is x Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Add 3 Multiply by 2 Add x Divide by 3 Subtract 1 The output is f(x)15 10 13 26 36 12 11 20 23 46 66 22 21 12 15 30 42 14 13 14 17 34 48 16 15 16 19 38 54 18 17 next 14

25. 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. Summary next © 2007 Herbert I. Gross

26. 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. next Summary © 2007 Herbert I. Gross