# Finding domains of functions

I'm doing college algebra homework and I always think that if something seems too easy, you're probably wrong, so I wanted to check. The domain of a function is all real numbers unless it creates a zero in the denomonator, yes? So for the following, the domain would be negative infinity to positive infinity? \begin{align} f(x)&=3x-18\\ \\ f(x)&=4x^2-3x+2 \\ \\ f(x)&=x^3-3x^2+5x+7 \\ \\ f(x)&=|3x|+17 \\ \\ f(x)&=x \end{align} There are some more that are square roots and at the bottom of fractions, so I know the answer to those will have numbers excluded, but it seems like this is a lot of examples where the answer would be "all real numbers" so I wanted to make sure I wasn't missing something. Thanks in advance people!

• If you can say that $f(whatever)$ has a definite value, then $whatever$ is in the domain of $f$. Is there any $whatever$ for which these functions do not have a value? – user2357112 supports Monica Apr 22 '14 at 16:47

Informally (and most probably that is enough in your situation) you are correct if you say that $\mathbb R$ is the domain of these functions. More careful would be the statement that $\mathbb R$ can serve as domain of these functions. This because you can also have a function prescribed by e.g. $x\mapsto 3x-18$ that has a subset like $(0,1)$ of $\mathbb R$ as domain. In that sense I would call $\mathbb R$ the maximal domain of these functions.
Your remark ending with "...unless it creates a zero in the denominator" is not correct. That is sufficient condition for not belonging to the domain, but not a necessary condition. You can have a function prescribed by $x\mapsto\sqrt{x}$. There is no denominator, but nevertheless all values $<0$ cannot belong to the domain.