Consider the function:
$f(x)=\frac{2x^4+7x^3-7x^2-3x+1}{x^3-x^2+x-1}.$
When looking for domain, the denominator $x^3-x^2+x-1$ can not be zero. So $f(x)$ is undefined when $x=1.$
But when simplifying given function I get:
$f(x)=\frac{2x^4+7x^3-7x^2-3x+1}{x^3-x^2+x-1}=\frac{(x-1)(2x-1)(x^2+4x-1)}{(x-1)(x^2+1)}=\frac{(2x-1)(x^2+4x-1)}{x^2+1}$.
Now domain seems to change, since $x^2+1>0$ for every $x \in \mathbb{R}$.
Does the domain change when simplifying given expression?
Also, when determinating the vertical asymptote, we look for domain. I was thinking that since $x^2+1>0$ for every $x \in \mathbb{R}$ that given function doesn't have any undefined points so vertical asymptote doesn't exist.
Or do I need to consider first domain and look for the limit of function as $x$ approaches $1^+$ or $1^-$?
Thank you!
Best Answer
It is $$f(x)=\frac{(x-1)(2x-1)(x^2+4x-1)}{(x-1)(x^2+1)}=\frac{(2x-1)(x^2-4x-1)}{x^2+1}$$ only if $$x\neq 1$$ There is a whole in this function at $x=1$