Domain of the function f(x) = frac{x - 3}{(x | vedclass

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Question

(B) For the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ to be defined,the expression inside the square root must be strictly positive and th

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ to be defined,the expression inside the square root must be strictly positive and the denominator must not be zero.$1$. The condition for the square root is $x^2 - 4 > 0$,which implies $x^2 > 4$,so $x \in (-\infty, -2) \cup (2, \infty)$.$2$. The condition for the denominator is $(x - 1)\sqrt{x^2 - 4} 
eq 0$. This means $x - 1 
eq 0$ (so $x 
eq 1$) and $\sqrt{x^2 - 4} 
eq 0$ (so $x 
eq \pm 2$).$3$. Combining these conditions,we need $x \in (-\infty, -2) \cup (2, \infty)$ and $x 
eq 1$. Since $1$ is not in the interval $(-\infty, -2) \cup (2, \infty)$,the domain remains $(-\infty, -2) \cup (2, \infty)$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

Domain of the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ is

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$y = f(x) = x^2$

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