Having the function $f(x) =\frac{x^3+3x^2}{2x^2+4x}$, why it is not the same to analyze $\frac{x^2+3x}{2x+4}$, if it verifies $\frac{x^3+3x^2}{2x^2+4x}=\frac{x^2+3x}{2x+4}$ ? In this case, the first one has only one root, while the second one has another one in $0$.
In general, can a function be simplified before being analyzed? (I mean, find roots, continuity, maxima, and minima...)
On
It is perfectly acceptable to cancel common factors of the numerator and denominator, provided that you retain the original domain.
Presumably, any point at which such factors vanish are not in the original domain, so division by zero is not a consideration.
For example, the real functions $f(x)= x^2/x$ and $g(x)=x$ are identical if they are regarded as maps $\mathbb R\setminus\{0\} \to\mathbb R$.
On
Hardy is fairly robust on this issue (and while possibly slightly at variance with current set-theoretic dogma, nevertheless quite consistent with the standard calculus interpretations):
The function $\frac{x^2-1}{x-1}$ has no value for $x=1$; for $x=1$, $\frac{x^2-1}{x-1}$ is strictly and absolutely meaningless. The fact that its limit for $x=1$ is $2$ is entirely irrelevant. The functions $\frac{x^2-1}{x-1}$ and $x+1$ are different functions. They are equal when $x$ is not equal to $1$. Similarly the function $y=\frac{x}{x}$ is $=1$ when $x \neq 0$ and undefined for $x=0$. To calculate $f(x)$ for $x=0$ we must put $x=0$ in the expression of $f(x)$ and perform the arithmetical operations which the form of the function prescribes, and this we cannot do in this case.
(Mathematical Gazette 1907, 4 pp. 13–14)
The function under analysis can generally be simplified. But note that when you simplify, canceling top and bottom of the same fraction is only valid if you are not dividing by zero.