# Why is the domain of $f(x)=\frac{2x^2}{x^4+1}$ all real numbers? It makes no sense.

The function $$f(x)=\frac{2x^2}{x^4+1}$$ has a domain of all real numbers. I cannot figure out exactly why. Shouldn't $x^4+1$ be incapable of equaling 0? I'm sure it is a very basic question, but it is kicking my ass.

• Yes, $x^4+1$ is never zero for $x$ real. That is why the domain is all of $\mathbb R$. Mar 15, 2015 at 20:03
• The domain means that plugging any value of $x$ gives you a well defined real number.And the co-domain means the value that is produced by the function Mar 15, 2015 at 20:04
• Co-domain, range, etc. are fuzzily defined, sadly, meaning one thing in one place, and others in other places. In particular, when we write $f:X\to Y$, in many places, the codomain is $Y$ whether or not $f$ is onto. @kingW3 Mar 15, 2015 at 20:08

The domain of a function $f$ is all values $x$ such that $f(x)$ is defined. In this context, "all values $x$" means "all real numbers $x$" and "is defined" means "is a real number".

Because the denominator of $\dfrac {2x^2}{x^4 + 1}$ is never zero, there is never a problem of $f(x)$ being undefined. Thus any real number $x$ has an appropriate $f(x)$, and that is why the domain is the entirety of $\mathbb R$.

• Okay, I get it now. Thanks! Mar 15, 2015 at 20:27

Yeah, its domain is the set of Real Numbers since $x^4+1$ cant be zero, since $x^4$ is always positive or zero. So there is no $x$ such that the rational function will be undefined. So the domain is $R$. For the basis,

Note: If $g(x)$ is a rational function of the form $\frac{n(x)}{d(x)}$ then the domain of $g(x)$ is the set of all real numbers except for those $x$ such that $d(x)=0$.