The natural domain of the real-valued functio | vedclass

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(D) The function is given by $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$.For the function to be defined in the set of real numbers,the expressions under

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$(-\infty, \infty) - (-1, 1)$

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(D) The function is given by $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$.For the function to be defined in the set of real numbers,the expressions under the square roots must be non-negative.For the first term $\sqrt{x^2 - 1}$,we require $x^2 - 1 \ge 0$,which implies $x^2 \ge 1$. This inequality holds when $x \in (-\infty, -1] \cup [1, \infty)$.For the second term $\sqrt{x^2 + 1}$,we require $x^2 + 1 \ge 0$. Since $x^2 \ge 0$ for all real $x$,$x^2 + 1 \ge 1$ is always true for all $x \in \mathbb{R}$.The domain of the function $f(x)$ is the intersection of the domains of the two terms.Thus,the domain is $(-\infty, -1] \cup [1, \infty)$,which can be written as $(-\infty, \infty) - (-1, 1)$.

The natural domain of the real-valued function defined by $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$ is

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