The natural domain of the real valued functio | vedclass

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(D) For the function $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$ to be defined,both square root terms must be non-negative.$1$. For $\sqrt{x^2 - 1}$,we r

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$x \in (-\infty, -1] \cup [1, \infty)$

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(D) For the function $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$ to be defined,both square root terms must be non-negative.$1$. For $\sqrt{x^2 - 1}$,we require $x^2 - 1 \ge 0$,which implies $x^2 \ge 1$. This holds when $x \le -1$ or $x \ge 1$.$2$. For $\sqrt{x^2 + 1}$,we require $x^2 + 1 \ge 0$,which is true for all real numbers $x \in \mathbb{R}$.Taking the intersection of these two conditions,the domain is $x \in (-\infty, -1] \cup [1, \infty)$,which can also be written as $(-\infty, \infty) - (-1, 1)$.

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0.25$	$0.5$	$1$	$1.5$	$2$
$y = \frac{1}{x}$	....	....	....	....	....	....	....	....	....

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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