Domain of the function f,given by f(x) = frac | vedclass

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Question

(B) For the function $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ to be defined,the expression inside the square root must be strictly greater than zero.

Vedclass · Accepted Answer

$(-\infty, 2) \cup (5, \infty)$

Vedclass · Answer

(B) For the function $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ to be defined,the expression inside the square root must be strictly greater than zero. $\Rightarrow (x - 2)(x - 5) > 0$ The critical points are $x = 2$ and $x = 5$. Testing the intervals $(-\infty, 2)$,$(2, 5)$,and $(5, \infty)$: For $x  0$. For $2 For $x > 5$,$(x - 2)(x - 5) > 0$. Thus,the domain is $x \in (-\infty, 2) \cup (5, \infty)$.

Domain of the function $f$,given by $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ is

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