Confusion regarding the domain of a function Let $f(x)=x^2$ and $g(x)=x$. What is the domain of $\frac{f(x)}{g(x)}$?
Evaluating $\frac{f(x)}{g(x)}$ gives us $x$. Does that mean that its domain is all real numbers?
If we evaluate the function at $x=0$, $\frac{f(0)}{g(0)}$, then $g(0)$ will gives us zero. Does that mean zero is not in the domain of $\frac{f(x)}{g(x)}$?
From what I understand, the domain of $x$ is the set of all real numbers, but the domain of $\frac{f(x)}{g(x)}$ is the set of all real numbers except zero. Am I right?
Edit: $f(x)=2x^2$. Sorry. I forgot to add the two. To make it less confusing, I'm just gonna remove the "2" in $2x$

The answer to your question is: It depends. Namely on the domain specified for $f$ and the domain specified for $g$.

Attention: A function is not fully specified as long as the domain of the function is not given.

At first you have to check the domain of the function $f$ and the domain of the function $g$. Although it seems natural that the domain is the largest possible set for which a function gives reasonable values, this has always to be clarified before doing some calculation.

The domain of $f/g$ is the intersection of the domain of $f$ and $g$ minus all points where $g(x)=0$.
Example 1:
  \begin{align*}
&f:\mathbb{R}\rightarrow\mathbb{R}&\qquad &g:\mathbb{R}\rightarrow\mathbb{R}\\
&f(x)=2x^2&\qquad &g(x)=x\\
\\
&\frac{f(x)}{g(x)}=2x&\qquad &x\in\mathbb{R}\setminus\{0\}
\end{align*}
Example 2:
  \begin{align*}
&f:\mathbb{R^+}\rightarrow\mathbb{R}&\qquad &g:\mathbb{R^+}\rightarrow\mathbb{R}\\
&f(x)=2x^2&\qquad &g(x)=x\\
\\
&\frac{f(x)}{g(x)}=2x&\qquad &x\in\mathbb{R}^+
\end{align*}
Example 3:
  \begin{align*}
&f:\{0\}\rightarrow\mathbb{R}&\qquad &g:\mathbb{R}\rightarrow\mathbb{R}\\
&f(x)=2x^2&\qquad &g(x)=x\\
\\
&\frac{f(x)}{g(x)} &\qquad&\text{ is not defined}
\end{align*}

