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ight) = 2\left( rac{2}{x^2 - 1}
ight) = rac{4}{x^2 - 1}.
Since $ x > 1 $, $ x^2 - 1 > 0 $, so $ f(x) > 0 $, and as $ x o 1^+ $, $ f(x) o \infty $, as $ x o \infty $, $ f(x) o 0^+ $. So $ R $ has no minimum, only an infimum of 0. But the problem likely intends for us to find the minimum under constraints — but none restrict. However, reconsider the original expression:

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