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

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Question

(B) For the function $f(x) = \sqrt{\frac{x-2}{3-x}}$ to be defined,the expression under the square root must be non-negative: $\frac{x-2}{3-x} \geq 0$

Vedclass · Accepted Answer

$[2, 3)$

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(B) For the function $f(x) = \sqrt{\frac{x-2}{3-x}}$ to be defined,the expression under the square root must be non-negative: $\frac{x-2}{3-x} \geq 0$. This implies that the numerator and denominator must have opposite signs or the numerator must be zero. Specifically,$x-2 \geq 0$ and $3-x > 0$. From $x-2 \geq 0$,we get $x \geq 2$. From $3-x > 0$,we get $x Combining these two conditions,we get $2 \leq x Thus,the domain of $f(x)$ is $[2, 3)$.

The domain of the real valued function $f(x) = \sqrt{\frac{x-2}{3-x}}$ is...

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