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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.enter image description here

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  • $\begingroup$ Grossly well defined means that whatever defines the function makes sense. For instance, in $\Bbb R$, the function $x\mapsto \sqrt{1-x^2}$ isn't well defined because, if $|x|>1$, you get the square root of a negative number. $\endgroup$ – Git Gud Jul 23 '13 at 13:02
  • $\begingroup$ Thank you for your feedback. $\endgroup$ – Dee Jul 23 '13 at 13:11
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    $\begingroup$ The word "singular" doesn't appear in the text you quote, so I'm not sure why you ask what "they" mean by it. $\endgroup$ – Chris Eagle Jul 23 '13 at 13:25
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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.

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  • $\begingroup$ Thanks, please note the function in question is f(x) = xe^x - 2 $\endgroup$ – Dee Jul 23 '13 at 13:27
  • $\begingroup$ See the last section. $\endgroup$ – AlexR Jul 23 '13 at 13:30

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