Question

The natural domain of the real valued function defined by
f(x)=root x square-1 + root x square+1 is
​

Answer 1

f(x) = √(x²−1) + √(x²+1)

So, the domain is given by the solution of the inequation :

x²−1 ≥ 0

(x+1)(x−1) ≥ 0

∴ x ∈ (−∞,−1] ∪ [1, +∞)

In the function f(x), (x²+1) ≥ 0 ∀ x ∈ ℝ

So, √(x²+1) ≥ 1