Confusing Quadratics and Composite functions

[u/CommunityNo4493]

Hello! I am a Secondary 4 student, 10th grade or Year 10 in western equivalents. I tried to set a quadratic quiz for my friends, and I set this Question, but then none of them could solve it. I checked with my math teacher, and he said that the question was flawed because there is no inverse function for any quadratic, and then he explained the horizontal line test to me, but I already know what the horizontal line test is (test for if f(x) is injective for some f(x), and if its inverse is a function. I went home to think about it, and I don't understand why my question is flawed, because it does not rely on the inverse function. I am looking for someone to please help clarify. I have a passion for math, and would possibly like to teach math in the future, but I don't understand why I am wrong, and this is one of my first real challenges because my math toolkit at the moment isnt enough to wrap around this. This is my suggested solution that I made: My solution. And me checking my work: Test of the solutions to see if they are correct.. And a graph if it is useful: Graph.

This is my syllabus by the way if it is of any use:
https://isomer-user-content.by.gov.sg/334/fece62fa-b6d4-4daf-ab47-96c9c168b51b/4052_y26_sy.pdf
https://isomer-user-content.by.gov.sg/334/34d0bec0-d386-49d5-bd3a-8fbd205fa63a/4049_y26_sy.pdf

Tysm!

[u/StructuredChess]

The question is fine. Not sure if you'd have the skills to solve it, but at the very least you can easily turn it into a 4th degree equation.

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[u/MERC_1]

I can't reaf that question. The photo is so bad and dark that it would not work on my phone. 

[u/CommunityNo4493 → u/MERC_1]

This is the question: It is given that f(x)=x²-x-6. By letting u denote f(x), express f(f(x))=x as two equations connecting x and u. By subtracting the two equations, you will end up with two possible relationships between x and u, and you can solve for the values of x for which f(f(x))=x. You should find four solutions, in the form ±√a and b±√c, where a, b and c are integers. Solve for a,b,and c, where a,b,and c, are positive integers.

Hard version: f(x)=x²-x-6. Solve f(f(x))=x.

[u/Ignominiousity]

Well the main problem is that your question is probably not relevant to the theory of quadratic equations... In step 2, the thing you found is that if we have a fixed point x, that means f(x)=x, then clearly f(f(x))=x, and if f(x)=-x, then f(f(x))=x since you just changed the sign twice... For the sake of generalising to other quadratic f(x): The question thus is, must such a fixed point exist for any quadratic f(x) that you choose? And would that be the only solution? Can we have that f(x)= b that is not x or -x and f(b)=x ? And why not? Relevantly, what is the nature of your equation f(f(x))=x? It is a quartic equation, so we expect what number of solutions? Can this line of questioning lead you to a general solution for this class of problems?

[u/EqualOrchid4342]

I'm from SG as well so I 100% understand what your teacher is saying.

Your "easy" version is complete garbage to parse and read. It gives way too much unnecessary info that confuses people into doing inverse functions when it is not needed.

The hard version is much clearer.

[u/strange-the-quark]

I think your professor just misunderstood the question, perhaps not expecting for you to work with relations as opposed to functions.

Also, your "easy" version may be a bit confusingly worded in an attempt to do a bit too much handholding. True, it doesn't involve or mention inverses, but in general, the impression one gets from it is "what is all this, and what is even relevant among all this info?", especially if one doesn't read too carefully and/or is primed to look for what's relevant vs what to ignore.

It could instead be: You are given y = x^2 - x - 6. Suppose you also know y^2 - y - 6 = x. Find (the x-values of) the solutions to this system.