Extract 7.6
Context: This is the beginning of the tutorial for students Abidul and Frances. They are discussing B7.
The Episode:
The tutor asks the students how they 'decided there was just one root' in part a (fig.6).
A1: I found the stationary points and there were two, so I said after the minimum the graph has to go up and after the maximum it has to go down and so it has to cut the graph somewhere because it tends to infinity...
The tutor nods in approval and asks Abidul what theorem she used in order to determine that because f goes to infinity as x tends to -¥, f must have a zero. The student is hesitant and whispers 'the Rolle's'. The tutor waits for a few seconds and then repeats the question to Frances. Frances replies it was the IVT and the tutor asks what does the Intermediate Value Theorem say.
F1: [after a pause] If f is continuous...then. And if f of...if b is greater than a, then f(b) is going to be...
A2: F(a) is less than 1 over...Then c...
T1: [laughing] I think between you, you've got a correct version. So we've got a c somewhere...where?
A3: In a,b. And it is f(c)= lamda...
The tutor completes the outline of the proof by pointing out that the IVT guarantees that f takes all the values between its maximum and -¥ and therefore it has to cut the x axis at one point.
In part b (fig.6) both students found that f has a min at 1, to the left of which it is strictly decreasing and to the right is strictly increasing. Therefore Imf=f((0,¥))=[e, +¥). They both have drawn the graph but cannot answer why Imf=[e, +¥). Abidul mumbles that she used the second derivative and continuity and that the function is monotone increasing. The tutor prompts them by saying that, from IVT they know that Imf must be an interval because f is continuous. The other thing they need to know, she continues, is that f is unbounded above. The tutor points at the graphs they drew and notices that they have drawn f as an unbounded function. She asks them why they 'sent off' the graph like this. Frances whispers that she drew f like that 'because as x gets very small...1/x...'. The tutor nods in agreement then stops Frances and completes the argument: 1/x®¥ and ex multiplies it with something big.
About the second part of b, Abidul says she didn't 'really know what to do' and Frances suggests I=(0,1). The tutor nods in approval. After a pause she responds to the tutor's request for an explanation with:
F2: Because you can tell from the graph! [the tutor asks why again and a pause follows] If...you reflect it on y...
The tutor asks her what theorem she used in order to guarantee the existence of the inverse function. The students are silent and the tutor states the Inverse Function Theorem and outlines its use for the particular function in question.
Return to Section 7(vi).Return to Appendices for Chapter 7.