Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a clarification to make with my notes. This is about functions. It defines the function

$$f(x) = \sqrt[5]{x+1} \space\text{from} \space x=(-1,\infty) \space \text{to} \space[0,\infty)$$

And it defines the inverse as

$$f^{-1}(x) = x^5 - 1 \space\text{from} \space x=[0,\infty) \space \text{to} \space(-1,\infty)$$

I have some questions:

Shouldn't the range of $f$ be $(0,\infty) $ instead of $[0,\infty) $? Either we change that or we say

$$f(x) \space \text{is from}\space x=[-1,\infty) \space \text{to} \space[0,\infty)\\ \text{and} \\ f^{-1}(x) \space \text{is from}\space x=[0,\infty) \space \text{to} \space[-1,\infty) $$

Please help.

UPDATE: Thank you for your help. I think there are 2 ways to do this: either the one above or:

$$f(x) \space \text{is from}\space x=(-1,\infty) \space \text{to} \space(0,\infty)\\ \text{and} \\ f^{-1}(x) \space \text{is from}\space x=(0,\infty) \space \text{to} \space(-1,\infty) $$

share|cite|improve this question
You're right. It's most likely a typo in the notes you've been reading. Shit happens. – Harald Hanche-Olsen Feb 24 '13 at 16:04
@GitGud i have made the edits, it was a typo. regarding your first point, I have 23 questions (incl. this one) with 21 accepted answers. That's less than 10% so please do not falsely accuse me. Thank you. – bryansis2010 Feb 24 '13 at 16:06
@bryansis2010 My bad, I mixed your questions with your answers.I apologize. I'll remove my comment. – Git Gud Feb 24 '13 at 16:08
@bryansis2010 As an alternative view, $f$ is a continuous function on an open interval, therefore it's range will be an open interval. – Git Gud Feb 24 '13 at 16:15
@GitGud no problems, could you just help me to check my update? Consider this one to have an accepted answer too. – bryansis2010 Feb 24 '13 at 16:17
up vote 3 down vote accepted

We always know that if $f:X\to Y$ be a one-one and onto and so its inverse exists; then : $$D_f=R_{f^{-1}}=X$$ and $$D_{f^{-1}}=R_f=Y$$

I have a counterexample to the above statement. The function $$y=x^{3}-1$$ is 1-1 and onto. Its domain and range are set of all real numbers. But its inverse $$y=(x+1)^{1/3}$$ has domain $$[-1,\infty)$$ and range $$[0,\infty).$$ How would you see your conclusion in this situation?

share|cite|improve this answer
I will take your answer because it goes back to the fundamentals that I learnt 9 years ago. With a bunch of notes like that, I'm glad the community helps. – bryansis2010 Feb 24 '13 at 16:15
@bryansis2010: Thanks. All of us are here just to make this puzzel complete. :-) – Babak S. Feb 24 '13 at 16:17
Hey, could you assist to check my question update to make sure I didn't make any schoolboy mistakes? – bryansis2010 Feb 24 '13 at 16:18
Why the downvote? – Git Gud Feb 24 '13 at 16:18
@BabakS. there's a typo in a subscript, you forgot $f^{-1}$. – Git Gud Feb 24 '13 at 16:19

The given function $f$ is increasing(1st derivative +ve) so it is one to one and onto.

Thus the inverse exists(One must always check the existence of inverse before talking abt it) and the inverse is given by the formula you have stated and the domains and the ranges are also the one you have mentioned(Basically if the inverse exists then the domain and range gets interchanged in the original function and the inverse.Quite easy to see if you draw the graph).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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