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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $D \supset [-1,1]$.

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.

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.