Domain and range of f(x) = frac{|x - 3|}{x - | vedclass

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(B) The function is defined as $f(x) = \frac{|x - 3|}{x - 3}$.For the function to be defined,the denominator must not be zero,so $x - 3 
eq 0$,which

Vedclass · Accepted Answer

$R - \{3\}, \{1, -1\}$

Vedclass · Answer

(B) The function is defined as $f(x) = \frac{|x - 3|}{x - 3}$.For the function to be defined,the denominator must not be zero,so $x - 3 
eq 0$,which implies $x 
eq 3$. Thus,the domain is $R - \{3\}$.Now,consider the two cases for the absolute value:Case $1$: If $x > 3$,then $|x - 3| = x - 3$. Therefore,$f(x) = \frac{x - 3}{x - 3} = 1$.Case $2$: If $x Thus,the range of the function is the set $\{1, -1\}$.

Domain and range of $f(x) = \frac{|x - 3|}{x - 3}$ are respectively

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