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

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Question

(C) For the function $f(x) = \sqrt{2 - x} - \frac{1}{\sqrt{9 - x^2}}$ to be defined:$1$. The expression under the square root in the numerator must be

Vedclass · Accepted Answer

$(-3, 2]$

Vedclass · Answer

(C) For the function $f(x) = \sqrt{2 - x} - \frac{1}{\sqrt{9 - x^2}}$ to be defined:$1$. The expression under the square root in the numerator must be non-negative: $2 - x \ge 0 \Rightarrow x \le 2$.$2$. The expression under the square root in the denominator must be strictly positive: $9 - x^2 > 0 \Rightarrow x^2 $3$. Combining these two conditions: $x \le 2$ $AND$ $-3 $4$. The intersection of these intervals 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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