The function f(x) = sqrt{1 - sqrt{1 - x^2}} | vedclass

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(D) For the function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$,the domain is determined by $1 - x^2 \ge 0$ (so $-1 \le x \le 1$) and $1 - \sqrt{1 - x^2} \ge

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All of the above

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(D) For the function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$,the domain is determined by $1 - x^2 \ge 0$ (so $-1 \le x \le 1$) and $1 - \sqrt{1 - x^2} \ge 0$ (which is always true for $x \in [-1, 1]$). Thus,the domain is $[-1, 1]$.To check differentiability at $x = 0$,we find the one-sided derivatives:$f'(0^+) = \lim_{h 	o 0^+} \frac{\sqrt{1 - \sqrt{1 - h^2}} - 0}{h} = \lim_{h 	o 0^+} \sqrt{\frac{1 - \sqrt{1 - h^2}}{h^2}} = \lim_{h 	o 0^+} \sqrt{\frac{1 - (1 - h^2)}{h^2(1 + \sqrt{1 - h^2})}} = \lim_{h 	o 0^+} \sqrt{\frac{h^2}{h^2(1 + \sqrt{1 - h^2})}} = \frac{1}{\sqrt{2}}$.Similarly,$f'(0^-) = -\frac{1}{\sqrt{2}}$.Since the left-hand derivative and right-hand derivative are finite but not equal,the function is continuous but not differentiable at $x = 0$. Therefore,all statements are correct.

The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

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