The domain of f(x) = sqrt{left(frac{1}{sqrt{x | vedclass

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(C) For the function $f(x) = \sqrt{\frac{1}{\sqrt{x}} - \sqrt{x+1}}$ to be defined,the following conditions must be met:$1$. The expression inside the

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$\left(0, \frac{\sqrt{5}-1}{2}\right]$

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(C) For the function $f(x) = \sqrt{\frac{1}{\sqrt{x}} - \sqrt{x+1}}$ to be defined,the following conditions must be met:$1$. The expression inside the square root must be non-negative: $\frac{1}{\sqrt{x}} - \sqrt{x+1} \geq 0$.$2$. The term $\sqrt{x}$ in the denominator requires $x > 0$.$3$. The term $\sqrt{x+1}$ requires $x+1 \geq 0$,which means $x \geq -1$.Combining these,we need $x > 0$.Now,solve the inequality: $\frac{1}{\sqrt{x}} \geq \sqrt{x+1}$.Since $x > 0$,we can square both sides: $\frac{1}{x} \geq x+1$.$1 \geq x^2 + x \implies x^2 + x - 1 \leq 0$.The roots of $x^2 + x - 1 = 0$ are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.Since the parabola opens upward,$x^2 + x - 1 \leq 0$ for $x \in \left[\frac{-1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right]$.Intersecting this with the condition $x > 0$,we get $x \in \left(0, \frac{\sqrt{5}-1}{2}\right]$.

The domain of $f(x) = \sqrt{\left(\frac{1}{\sqrt{x}} - \sqrt{x+1}\right)}$ is

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