Respuesta :
Assuming you're asking "for which values of [tex]n[/tex] the function [tex]x^n[/tex] has an inverse that is a function", the answer is "all the odd exponents [tex]n[/tex]".
Infact, if [tex]n[/tex] is even, you have that
[tex]x^n=(-x)^n \quad \forall x \in \mathbb{R}[/tex]
and so [tex]f(x)=x^n[/tex] is not injective, and thus not invertible
On the other hand, if [tex]n[/tex] is odd, we have:
- [tex]\lim_{x\to\pm\infty}x^n=\pm\infty[/tex]
- [tex]x^n[/tex] is continuous.
- The first two points tell us that the function is surjective.
- Moreover, the derivative is [tex]f'(x)=nx^{n-1}[/tex]. Since [tex]n-1[/tex] is even, we have [tex]f'(x)>0[/tex], thus the function is always increasing, and so the function is also injective.
- Injective and surjective means bijective, and the function can be inverted.
