How do you find the domain and range for $f(x)= (x-3)/( x^2- 9)$ ?

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Answer 1 · 2016-02-06T00:11:53

$(x-3)/(x^2-9)$

$=(x-3)/((x-3)(x+3)), x!=3, x!=-3$

$=1/(x+3), x!=3, x!=-3$

The domain will be $RR$ \ ${-3,3}$

Remember that $x!=3$ , there $1/(3+3)=1/6$ is not valid

therefore the range will be $RR$ \ ${1/6}$

How do you find the domain and range for f(x)= (x-3)/( x^2- 9)f(x)= (x-3)/( x^2- 9)?