Answer:
1
Step-by-step explanation:
Assuming you're asking "for which values of
the function
has an inverse that is a function", the answer is "all the odd exponents
".
Infact, if
is even, you have that
![x^n=(-x)^n \quad \forall x \in \mathbb{R}](https://tex.z-dn.net/?f=x%5En%3D%28-x%29%5En%20%5Cquad%20%5Cforall%20x%20%5Cin%20%5Cmathbb%7BR%7D)
and so
is not injective, and thus not invertible
On the other hand, if
is odd, we have:
![\lim_{x\to\pm\infty}x^n=\pm\infty](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%5Cpm%5Cinfty%7Dx%5En%3D%5Cpm%5Cinfty)
is continuous.- The first two points tell us that the function is surjective.
- Moreover, the derivative is
. Since
is even, we have
, thus the function is always increasing, and so the function is also injective. - Injective and surjective means bijective, and the function can be inverted.
Answer:
Step-by-step explanation:
![\frac{6}{\sqrt{2} - \sqrt{3} } * \frac{\sqrt{2} + \sqrt{3} }{\sqrt{2} + \sqrt{3} }](https://tex.z-dn.net/?f=%5Cfrac%7B6%7D%7B%5Csqrt%7B2%7D%20-%20%5Csqrt%7B3%7D%20%7D%20%20%2A%20%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%20%5Csqrt%7B3%7D%20%7D%7B%5Csqrt%7B2%7D%20%20%2B%20%5Csqrt%7B3%7D%20%7D)
![\frac{6\sqrt{2} + 6\sqrt{3} }{2-3}](https://tex.z-dn.net/?f=%5Cfrac%7B6%5Csqrt%7B2%7D%20%2B%206%5Csqrt%7B3%7D%20%7D%7B2-3%7D)
![-6(\sqrt{2} + \sqrt{3} )](https://tex.z-dn.net/?f=-6%28%5Csqrt%7B2%7D%20%2B%20%5Csqrt%7B3%7D%20%29)
b
Step-by-step explanation:
because it will be right always because if you mutiply all of them then you get your right answer