Okay, let's see. I need to find f inverse of f inverse of 6. Hmm. So, first, maybe I should figure out what f inverse is. Since f is a function and f inverse is its inverse, that means if f(a) = b, then f inverse(b) = a. Right? So, to find f inverse, I can just swap the inputs and outputs of the original function f.

Given that f(1) = 2, f(2) = 6, and f(3) = 5. Let me write those down as ordered pairs. So, f is defined by the points (1, 2), (2, 6), and (3, 5). Therefore, the inverse function f inverse would have the points (2, 1), (6, 2), and (5, 3). Let me confirm that. If f takes 1 to 2, then f inverse should take 2 back to 1. Similarly, f takes 2 to 6, so f inverse takes 6 to 2. And f takes 3 to 5, so f inverse takes 5 to 3. Yeah, that seems right.

So, f inverse is a function with domain {2, 6, 5} and codomain {1, 2, 3}. Now, the problem is asking for f inverse of f inverse of 6. Let's break that down. First, compute f inverse of 6, then take f inverse of that result.

Looking at the inverse function, f inverse(6) would be the x-value such that f(x) = 6. From the original function f, f(2) = 6. So, f inverse(6) is 2. Then, we need to find f inverse of 2. From the inverse function, f inverse(2) is the x-value where f(x) = 2. From the original function, f(1) = 2. Therefore, f inverse(2) is 1. So, putting it all together, f inverse of f inverse of 6 is 1.

Wait, let me double-check. Let's retrace. f inverse is the inverse, so f inverse(6) is the input to f that gives 6. Since f(2) = 6, f inverse(6) = 2. Then, f inverse(2) is the input to f that gives 2. Since f(1) = 2, f inverse(2) = 1. So yes, the result is 1. That seems straightforward.

Alternatively, maybe I can think of this in terms of composing functions. f inverse composed with f inverse is f inverse squared, but maybe that's complicating things. Alternatively, since f is a bijection (assuming it's invertible), then applying the inverse twice would cycle through the mappings. But in this case, since we only have three points, it's a finite function, so maybe the cycle is different. But in any case, step by step, the answer seems to be 1.

Just to ensure there's no mistake, let's verify each step again. Starting with 6, apply f inverse: the original function f maps 2 to 6, so f inverse maps 6 back to 2. Then, applying f inverse again to 2: the original function f maps 1 to 2, so f inverse maps 2 back to 1. So, yes, 6 goes to 2, then 2 goes to 1. Therefore, the final answer is 1. I think that's correct. I don't see any errors in the reasoning.
Therefore, $f^{-1}(f^{-1}(6))$ is $\boxed{1}$.
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