Finding the domain of an inverse function

Question

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:

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

Answer 1

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?

Answer 2

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).