f (x) = \( \dfrac{1}{\sqrt{x^2-1}}\)

We know the square of a real number is never negative.

f (x) takes real values only when x^{2} – 1 ≥ 0

x^{2} – 1^{2} ≥ 0

(x + 1) (x – 1) ≥ 0

x ≤ –1 or x ≥ 1

x ∈ (–∞, –1] ∪ [1, ∞)

In addition, f (x) is also undefined when x^{2} – 1 = 0 because denominator will be zero and the result will be indeterminate.

x^{2} – 1 = 0 ⇒ x = ± 1

So, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}

x ∈ (–∞, –1) ∪ (1, ∞)

Domain (f) = (–∞, –1) ∪ (1, ∞)

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