Well defined, continuous and singular Can you explain what they mean when a function is well defined, continuous and singular? I know a function is continuous when you look at the right and left hand limits and both conclude to the same number.
Am I right when I say option 5 is false? See attached picture.
 A: well defined:
The definition is not dependent on an individual representant. i.E. for
$$f: \mathbb{R}/2\pi\mathbb{Z} \to [0, 2\pi), \qquad [x] \mapsto x \ ({\rm Mod}\ 2\pi)$$
is well-defined, while
$$f: \mathbb{R}/2\pi\mathbb{Z} \to [0, 2\pi), \qquad [x] \mapsto x$$
is not, as in $\mathbb{R}/2\pi\mathbb{Z} 0 = 2\pi = 4\pi$, but $f([0]) = 0 \neq 2\pi = f(\underbrace{[2\pi]}_{=[0]})$.
Another requirement is, that for $f: D\to V$ holds $f(D) \subset V$, i.e. no values out of the scope are assigned.
$$ \sin: \mathbb{R} \to D$$
is only well-defined for $D \supset [-1,1]$.

Continuous:
There are multiple definitions, the most elementary definition is:
$$ f \text{ is continuous} :\Leftrightarrow f^{-1}(A) \text{ is open for every } A \text{ open} $$
This only requires a Topology.

Singularity:
A point in the definition set, for wich the function assignment is not well-defined
$$ f: \mathbb{R} \to \mathbb{R}, \qquad x\mapsto \frac{1}{x} $$
is singular in $x=0$.

Applied to your function: $f$ is clearly well-defined and continuous, so compute the derivative:
$$f'(x) = xe^x + e^x $$
as a composition of two well-defined and continuous functions, $f'$ is also well-defined and continuous. None of the statements are false.
