Domain of the real valued function f(x) = fra | vedclass

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(B) The function $f(x) = \frac{x+2}{9-x^{2}}$ is defined for all real values of $x$ except where the denominator is zero. Setting the denominator to z

Vedclass · Accepted Answer

$R - \{-3, 3\}$

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(B) The function $f(x) = \frac{x+2}{9-x^{2}}$ is defined for all real values of $x$ except where the denominator is zero. Setting the denominator to zero: $9 - x^{2} = 0$. This implies $x^{2} = 9$,so $x = \pm 3$. Thus,the function is undefined at $x = 3$ and $x = -3$. Therefore,the domain of the function is the set of all real numbers except $\{-3, 3\}$,which is written as $R - \{-3, 3\}$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

Domain of the real valued function $f(x) = \frac{x+2}{9-x^{2}}$ is

Similar Questions

Consider the following lists.
$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=|[x]|, x \in R$ $2$. $Z$
$C$. $h(x)=|x-[x]|, x \in R$ $3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$ $4$. $[0, 1)$
$5$. $\{-1, 1\}$

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