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

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(A) For the function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ to be defined,the expressions under the square roots must be non-negative: $1$. $x^2 - 144 \

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$[-15, -12] \cup [12, 15]$

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(A) For the function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ to be defined,the expressions under the square roots must be non-negative: $1$. $x^2 - 144 \geq 0$ $\Rightarrow x^2 \geq 144$ $\Rightarrow |x| \geq 12$. $2$. $9 - \sqrt{x^2 - 144} \geq 0 \Rightarrow 9 \geq \sqrt{x^2 - 144}$. Squaring both sides,we get $81 \geq x^2 - 144$ $\Rightarrow x^2 \leq 225$ $\Rightarrow |x| \leq 15$. Combining both conditions,we have $12 \leq |x| \leq 15$. This implies $x \in [-15, -12] \cup [12, 15]$.

The domain of the real valued function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ is

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