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f(x)≤2(1+x)∀x∈[1,∞)xf(x+1)=(f(x))2−1∀x∈[1,∞)\begin{aligned} f(x) & \le 2(1+x) & \forall x \in [1, \infty) \\ x f (x+1) & = (f(x))^2 - 1 & \forall x \in [ 1,\infty ) \end{aligned} f(x)xf(x+1)≤2(1+x)=(f(x))2−1∀x∈[1,∞)∀x∈[1,∞)
A function f:[1,∞)→[1,∞)f:[1, \infty) \rightarrow [1, \infty) f:[1,∞)→[1,∞) satisfy the conditions above. Find f(2016)f(2016)f(2016).
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