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Given the function
$$y=f(x)=\frac{x^2-9}{x-3},$$
I'd like to find its range.
I tried to find the range by first expressing $x$ in terms of $y$, but then I realized it's not possible to do so in this case. In what other way can I find the range of this function?
2 Answers
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Hint: Simplify $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}$$ if $x\neq 3$
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As has been pointed out $y=\frac{x^2-9}{x-3}=x+3$ when $x\ne 3$. Since $3$ is not in the domain, that means that $3+3=6$ is not in the range. So the range is all real numbers except for $6$. In interval notation, $y\in (-\infty, 6) \cup (6, \infty)$
