The domain of the real valued function f(x) = | vedclass

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(B) For the function $f(x) = \log_{\sqrt{2}}(\sqrt{x^2+x} + \sqrt{x^2-x})$ to be defined, the argument of the logarithm must be positive and the squar

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

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(B) For the function $f(x) = \log_{\sqrt{2}}(\sqrt{x^2+x} + \sqrt{x^2-x})$ to be defined, the argument of the logarithm must be positive and the square roots must be defined.$1$. For $\sqrt{x^2+x}$ to be defined, $x^2+x \ge 0 \implies x(x+1) \ge 0$. This gives $x \in (-\infty, -1] \cup [0, \infty)$.$2$. For $\sqrt{x^2-x}$ to be defined, $x^2-x \ge 0 \implies x(x-1) \ge 0$. This gives $x \in (-\infty, 0] \cup [1, \infty)$.$3$. The intersection of these two conditions is $x \in (-\infty, -1] \cup [1, \infty) \cup \{0\}$.$4$. Additionally, the argument $\sqrt{x^2+x} + \sqrt{x^2-x} > 0$. At $x=0$, the expression becomes $\sqrt{0} + \sqrt{0} = 0$, and $\log_{\sqrt{2}}(0)$ is undefined.Thus, we exclude $x=0$.The domain is $(-\infty, -1] \cup [1, \infty)$.

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

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