To solve functional equations where f(x) is not explicitly defined, substitute x with 1-x to create a system of equations, then eliminate the unknown function terms algebraically to find the explicit form of f(x). For the equation f(x) + 2f(1-x) = 3x², substituting x with 1-x gives f(1-x) + 2f(x) = 3(1-x)²; multiplying the second equation by 2 and subtracting the first yields 3f(x) = 6(1-x)² - 3x², so f(x) = 2(1-x)² - x², and f(2) = -2.
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A TMUA-Style Functional Equation ProblemHinzugefügt:
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?
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.
That could be nice.
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.
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.
So, this will be just be f of u = 3 * x squared, which is 1 - u squared.
So, this is true for all real numbers x and this is true for all real numbers u.
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.
And now what we can do is if we look at these two things here, it looks almost like simultaneous equations.
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.
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.
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