A TMUA-Style Functional Equation Problem

Ajouté : 2026-05-30

[00:00:00]Today, I'm going to be solving a very nice TMUA style problem. Let's have a look at this one. A function f from the reals to the reals, that's what this notation means. It means the domain is the reals and the set of outputs is a subset of the reals satisfies for all real x, f of x plus 2 * f of 1 - x = 3x squared. What is the value of f of 2?

[00:00:21]So, this is what we would call an implicit equation for f. We don't have f of x explicitly, normally we do, but in this problem we don't have f of x explicitly, we have this weird function with multiple f's in and we want to work out the value of f of 2. Pause the video now and give this problem a go for yourself. I'm going to dive right into a solution. There's a couple of natural things that you may want to do with these sorts of problems. You may want to start by plugging in values of x. You could plug in x is 0 is normally a nice one, especially because this right hand side will become 0. And in this case, x is 1. And you can maybe plug in some more values, maybe x is a half here cuz then you put a half and a half there.

[00:00:55]That could be nice.

[00:00:56]But the trick here, or one of the ways that we can solve this problem, is actually by noticing that if I replace x with 1 - x, x turns into 1 - x and 1 - x turns into x. So, just to make this a bit more clear, if I say let x = 1 - I'll use a different letter, u, then this is going to become f of u + 2 f of 1 - x will just be u. Sorry, I've messed this up.

[00:01:23]f of 1 - u, so x is 1 - u. So, f of x is f of 1 - u + 2 lots of f of 1 - x. Well, if x is 1 - u, that means u is 1 - x.

[00:01:33]So, this will be just be f of u = 3 * x squared, which is 1 - u squared.

[00:01:40]So, this is true for all real numbers x and this is true for all real numbers u.

[00:01:46]But there's nothing really special about the letter u. I can just replace the letters u here with x. Since this is true for all values of u, in particular it's going to be true when u = x. It's going to be true to say that f of 1 - x + 2 f of x = 3 1 - x squared like so for all x.

[00:02:04]And now what we can do is if we look at these two things here, it looks almost like simultaneous equations.

[00:02:10]And if we eliminate the f of 1 - x's from the here, we're just going to get an explicit equation for f of x from which we can then work out the value of f of 2.

[00:02:18]So what I'm going to do is take this second equation and multiply both sides by 2. So 2 f of 1 - x + 4 f of x = 6 lots of 1 - x squared and I'll call this equation equation 2 and this equation 1. And if I do equation 2 - equation 1 the f of 1 - x's will cancel and I'll be left with 4 f of x - f of x so 3 f of x = 6 6 lots of 1 - x squared - then 3 x squared like so. And if I just divide through by 3, I get f of x = 2 * 1 - x squared - x squared. And we want to know f of 2. So we've got f of x is this thing here. I mean I could expand this and simplify but there's no need. I can just plug in 2 into this and I get 2 lots of 1 - 2 squared - 2 squared. 1 - 2 is -1 all squared is 1 * 2 is 2 - 4 gives us -2. And so the answer here is -2.