Fixations and Madness

Our sixth and final post on the 2017 VCE exam madness is on some recurring nonsense in Mathematical Methods. The post will be relatively brief, since a proper critique of every instance of the nonsense would be painfully long, and since we’ve said it all before.

The mathematical problem concerns, for a given function f, finding the solutions to the equation

    \[\boldsymbol{(1)\qquad\qquad f(x) \ = \ f^{-1}(x)\,.}\]

This problem appeared, in various contexts, on last month’s Exam 2 in 2017 (Section B, Questions 4(c) and 4(i)), on the Northern Hemisphere Exam 1 in 2017 (Questions 8(b) and 8(c)), on Exam 2 in 2011 (Section 2, Question 3(c)(ii)), and on Exam 2 in 2010 (Section 2, Question 1(a)(iii)).

Unfortunately, the technique presented in the three Examiners’ Reports for solving equation (1) is fundamentally wrong. (The Reports are here, here and here.) In synch with this wrongness, the standard textbook considers four misleading examples, and its treatment of the examples is infused with wrongness (Chapter 1F). It’s a safe bet that the forthcoming Report on the 2017 Methods Exam 2 will be plenty wrong.

What is the promoted technique? It is to ignore the difficult equation above, and to solve instead the presumably simpler equation

    \[ \boldsymbol{(2) \qquad\qquad  f(x) \ = \  x\,,}\]

or perhaps the equation

    \[\boldsymbol{(2)' \qquad\qquad f^{-1}(x)\ = \ x \,.}\]

Which is wrong.

It is simply not valid to assume that either equation (2) or (2)’ is equivalent to (1). Yes, as long as the inverse of f exists then equation (2)’ is equivalent to equation (2): a solution x to (2)’ will also be a solution to (2), and vice versa. And, yes, then any solution to (2) and (2)’ will also be a solution to (1). The converse, however, is in general false: a solution to (1) need not be a solution to (2) or (2)’.

It is easy to come up with functions illustrating this, or think about the graph above, or look here.

OK, the VCAA might argue that the exams (and, except for a couple of up-in-the-attic exercises, the textbook) are always concerned with functions for which solving (2) or (2)’ happens to suffice, so what’s the problem? The problem is that this argument would be idiotic.

Suppose that we taught students that roots of polynomials are always integers, instructed the students to only check for integer solutions, and then carefully arranged for the students to only encounter polynomials with integer solutions. Clearly, that would be mathematical and pedagogical crap. The treatment of equation (1) in Methods exams, and the close to universal treatment in Methods more generally, is identical.

OK, the VCAA might continue to argue that the students have their (stupifying) CAS machines at hand, and that the graphs of the particular functions under consideration make clear that solving (2) or (2)’ suffices. There would then be three responses:

(i) No one tests whether Methods students do anything like a graphical check, or anything whatsoever.

(ii) Hardly any Methods students do do anything. The overwhelming majority of students treat equations (1), (2) and (2)’ as automatically equivalent, and they have been given explicit license by the Examiners’ Reports to do so. Teachers know this and the VCAA knows this, and any claim otherwise is a blatant lie. And, for any reader still in doubt about what Methods students actually do, here’s a thought experiment: imagine the 2018 Methods exam requires students to solve equation (1) for the function f(x) = (x-2)/(x-1), and then imagine the consequences.

(iii) Even if students were implicitly or explicitly arguing from CAS graphics, “Look at the picture” is an absurdly impoverished way to think about or to teach mathematics, or pretty much anything. The power of mathematics is to be able take the intuition and to either demonstrate what appears to be true, or demonstrate that the intuition is misleading. Wise people are wary of the treachery of images; the VCAA, alas, promotes it.

The real irony and idiocy of this situation is that, with natural conditions on the function f, equation (1) is equivalent to equations (2) and (2)’, and that it is well within reach of Methods students to prove this. If, for example, f is a strictly increasing function then it can readily be proved that the three equations are equivalent. Working through and applying such results would make for excellent lessons and excellent exam questions.

Instead, what we have is crap. Every year, year after year, thousands of Methods students are being taught and are being tested on mathematical crap.

23 Replies to “Fixations and Madness”

  1. Nice analysis, as always, Marty.

    I think we can depend on the Examiner’s Report to NOT mention any of this, or any of what I’m about to add. I’d like to add two things (I could add a whole lot more but life is short and opportunities should be left for others):

    1. There’s a very simple proof (totally within the scope of the Maths Methods Study Design) that if f is a strictly increasing function then the points of intersection of y = f(x) and y = f^-1(x) lie on the line y = x:

    Assume a point (a, b) that is an intersection point of y = f(x) and y = f^-1(x).
    Then f(a) = b and f^-1(a) = b => a = f(b).
    Case 1: Let a < b. Then f(b) b. Then f(b) > f(a) therefore f is not increasing on [b, a] which is a contradiction since f is a strictly increasing function.
    Therefore a = b and so the intersection point lies on y = x.

    (a) Cases 1 and 2 are symmetric and so either one is sufficient to conclude a = b.
    (b) *Strictly* increasing rather than increasing is required to ensure existence of the inverse function.

    This proof would make a nice context to and add relevance for teaching strictly increasing functions – at the moment the idea of strictly increasing functions is in the Study Design for no obvious reason. Just another bit of piecemeal added material.

    2. Part (i) (ii) provides a nice opportunity for applying the above theorem. By starting with f(x) = x the conjecture that the coordinates of the second intersection point approach (-2, -2) can be *proved*. From which it can be *proved* that the limiting bounded area is 4, rather than looking at some pictures and then giving a *conjectured* answer. A lost opportunity for VCAA. The proof is left as a simple exercise.

    The question is only worth 1 mark, so finding an answer by fair means or foul is encouraged here. This is an total indictment on VCAA , which consistently gets on its high horse and says that correct answers obtained by incorrect mathematics receive zero. While I agree with the sentiment, it’s totally sanctimonious and hypocritical when VCAA is more or less forcing students to answer exam questions simply by looking at some pictures and then making a conjecture. That is, forcing students to *incorrectly* answer the question. It’s incorrect because conjecture without *proof* is NOT correct mathematics.

    1. “…correct answers obtained by incorrect Mathematics receive zero.”

      Is this statement (presumably a copy-and-paste from an official VCAA publication somewhere) not itself a contradiction of the method mark/answer mark system by which 99% of marks are awarded in Methods papers? The extra 1% is a margin of error because I have seen a reasoning mark in a marking scheme recently but it may have been a HSC or even IB paper (the question was quite nice, so balance of probability dictates it probably wasn’t VCE)

      1. Thanks Number 8. I’m not sure if John was quoting or being rhetorical. Either way, what constitutes acceptable reasoning in Methods is anyone’s guess.

        1. It’s not a direct quote but it wouldn’t be hard to go to the reports and find this sort of statement. And who really knows how the marking scheme operates (accept the assessors who are sworn to confidentiality), it’s all top hush ….

          BY the way, my proof keeps publishing wrong, I give up. I’m sure readers can figure out what’s meant.

    2. I should clarify: What’s not correct mathematics is conjecture that is not clearly stated as conjecture. The Goldbach conjecture is good (great) mathematics, but I don’t think any competent mathematician would consider that being tested up to 400,000,000,000,000 constitutes a proof. But this is the sort of ‘proof’ that VCAA is encouraging with these sort of questions ….

      1. Wonderful! Love it!

        A bit like the “3 is a prime, 5 is a prime, 7 is a prime… all odd numbers are prime.” type of proof.

        That could be a whole new blog post Marty: “When is a proof not a proof?” Answer: when it is on a VCAA marking scheme.

        1. Yes. John, I think this a very important point, more generally on Methods. Whatever students are doing in Methods, and whatever its value, it’s not mathematical reasoning. It’s not establishing the truth of anything.

  2. And yet, if f is invertible (i.e. if f^(-1) exists), the simple solution of “solve f^(2)(x) = x” is a perfectly valid way to find all solutions to “f(x) = f^(-1)(x)”. Why *that* isn’t taught, I have no idea.

    1. Thanks, Eddie. Yes, for invertible functions the equation f2(x) = x is equivalent to (1), and is potentially much easier to solve. And, of course the iteration of a function is much more natural.

      Why isn’t it taught? I suspect the equations have no intrinsic interest for the VCAA examiners. It appears they’re more concerned with the pretty picture made by graphing f-1 and f together, than they are with actual function behaviour.

  3. It’s also sad to see the number of copycat questions appearing in school assessments and commercial trial exams that propagate and further entrench this method of solution. As someone who is often vetting this stuff, I am consistently having to alert the writer that such a method needs to be justified with the prior observation that either f or f^-1 is a strictly increasing function.

      1. Indeed. Unfortunately it’s the ‘Monkey see, monkey do’ mentality. Not helped when some commercial organisations (NOT MAV) have such poor insight that they demand their trial exam writers more or less copy the previous year’s exam.

      2. Yep, in a pandemic year, it’s become even clearer that some viruses just don’t get eliminated.

        The \displaystyle f(x) = f^{-1}(x) virus refuses to go away. There’s been another outbreak – this time in a 2020 Methods Trial Exam 1 sold by the Mouse AdenoVirus (how fitting in a pandemic year).

        The solution consists of several lines, the first one being:

        Solve f(x) = x 1M

        So 1 mark for a mathematically diseased statement – no mathematical basis for being valid. Unfortunately, this latest hotspot will no create new clusters of this sort of viral shit.

        The only vaccine I can think of is to insist on a refund (or re-write) of this exam on the grounds that the product is not 100% fit for purpose. And then maintain a suitable social (and commercial) distance from then on.

        As an aside, who thinks that \displaystyle \int \frac{f'(x)}{f(x)} dx = \ln ( f(x) ) + c is part of the *Maths Methods* course and therefore examinable? I’m very interested in what people think – those of you who have seen the 2011 Mouse AdenoVirus Trial Methods Exam 1 and the 2020 Mouse AdenoVirus Trial Methods Exam 1 (both written by the same person) will know why I’m interested …

  4. Just to prove that old posts don’t die, I have a bit more to add here!

    I’ve recently updated a detailed solution I wrote for the 2017 NHT Exam 1 Q8 – in particular I added a detailed discussion on solving the harder equations \displaystyle f(x) = f^{-1}(x) and \displaystyle g(x) = g^{-1}(x) and the importance of testing for extraneous solutions. In doing so, I realised that there’s another significant point (I think) to be made regarding part (c).

    When solving things like \displaystyle 2 \sqrt{2x+c}-2= (x+1)^2 -c you square both sides and in doing so typically introduce extraneous solutions. For this equation the potential solutions that require testing are \displaystyle x = \pm \sqrt{c-1} and \displaystyle x = \pm \sqrt{c-5} - 2.

    Potential solutions are tested by substituting them into the original equation and seeing whether they satisfy it or not.

    However, the Examination Report encourages students to test solutions by checking whether or not they lie in the domains of g and \displaystyle g^{-1}. This is OK when testing solutions arising from \displaystyle (x+1)^2 -c = 2x because there are no extraneous solutions – you’re simply applying a restriction on \displaystyle x = \pm \sqrt{c-1}. But it’s definitely NOT OK when testing solutions arising from \displaystyle 2 \sqrt{2x+c}-2= (x+1)^2 -c after you square both sides.

    For example, imposing domain restrictions leads to accepting \displaystyle x = \sqrt{c-5} - 2 as a solution when c \geq 6. But when it’s actually tested by substitution into \displaystyle 2 \sqrt{2x+c}-2= (x+1)^2 -c (a non-trivial calculation) you find that it does NOT work for any value of c – it’s an extraneous solution.

    You have to first test for extraneous solutions by substituting into the original equation, and then impose the domain restrictions on the solutions you decide to keep.

    So students who try and solve \displaystyle 2 \sqrt{2x+c}-2= (x+1)^2 -c get hit with a double whammy of non-trivial difficulties. VCAA is as oblivious to this second technicality as the first – VCAA is encouraging incorrect mathematics on several different fronts with this question, not just for encouraging exclusive use of technique that is not always valid.

    The question is actually a very rich one – it contains a lot of good mathematics when done properly. A shame that VCAA doesn’t do things properly.

    (As a minor point, part (c) was broken up into two parts each worth 2 marks, but it’s hard to see how to answer part (ii) without having done all of the necessary calculations in part (i), in which case part (ii) is simply 1 mark for an answer. So in my opinion, part (c) should have been a 3 + 1 mark split. And of course, you could easily argue that part (c) should have been worth more than 4 marks if it’s to be answered properly. And you could easily argue it should have been written as:

    c) Find the value(s) of c for which solutions to \displaystyle g(x) = g^{-1}(x) exist.
    3 marks

    d) Hence state the value(s) of c for which there is exactly one solution.
    1 mark)

    1. Or alternatively, impose domain restrictions and then test for extraneous solutions. The point is that both ‘tests’ are required if f(x) = f^{-1}(x) is solved and this point is completely ignored/overlooked/misunderstood by VCAA.

  5. What is the purpose asking students to solve f(x) = f^{-1}(x) in a short space of time?

    What is it that the examiners are trying to assess?

    1. Good question, Terry. I think the original idea, lost in the mists of time, was to be teaching and testing the symmetry of f and f-inv around y = x, and if properly taught that could work. But, as you noted before your edit, the question has ossified into a ritualistic, meaningless and unvalidated “trick”.

  6. Re: The appearance of this shit on the VCAA 2011 Exam 2 Question 3 (c) (ii). It also appears in part (d)(ii). There are VCAA produced solutions (in the form of a Wolfram notebook) where g(x) = x is used WITHOUT ANY JUSTIFICATION to solve part (d) (ii). The only practical way of solving part (d) (ii) is to use g(x) = x.

    1. The only practical way using a CAS calculator, that is.
      Mathematica can solve g(x)=g^{-1}(x) so there was no need (apart from efficiency of coding) to use the ‘short-cut’. Nevertheless, the short-cut was used, and used in a completely ignorant way. As we’ve learnt to expect from VCAA.

  7. Thanks, John. It appears to be a myth that won’t die. Nonetheless, I think VCAA may be becoming a little more circumspect with these problems. I don’t have time to find it now, but my memory was that on the recent exam questions of this type, the functions were obvious enough in some manner that going straight to f(x) = x was not so egregious (although the obviousness to do this should still have been noted.)

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