The domain of the function f(x) = sqrt{2 - x} | vedclass

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Question

(C) For the function to be defined,the expressions inside the square roots must satisfy specific conditions:$(i)$ For $\sqrt{2 - x}$ to be defined,$2

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$(-3, 2]$

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(C) For the function to be defined,the expressions inside the square roots must satisfy specific conditions:$(i)$ For $\sqrt{2 - x}$ to be defined,$2 - x \ge 0$,which implies $x \le 2$.(ii) For $\frac{1}{\sqrt{9 - x^2}}$ to be defined,the denominator must be non-zero and the value inside the square root must be positive: $9 - x^2 > 0$.This implies $x^2 To find the domain,we take the intersection of both conditions: $x \le 2$ $AND$ $-3 The intersection of $(-\infty, 2]$ and $(-3, 3)$ is $(-3, 2]$.Therefore,the domain is $(-3, 2]$.

The domain of the function $f(x) = \sqrt{2 - x} - \frac{1}{\sqrt{9 - x^2}}$ is

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