I think since the function is not defined for $x \lt 1$ so this function is not invertible. I had also proved that this is an one to one function but is this an onto function too?
Asked
Active
Viewed 100 times
1
-
3It is an invertible function from $(1,\infty)$ to $(1,\infty)$. – Kavi Rama Murthy Jul 19 '21 at 06:24
-
@Alan in the title it says its not – Vivaan Daga Jul 19 '21 at 06:35
-
The domain or range of a function can be $anything$. If the domain and range of $f $ are each $(1,\infty)$ then $f$ is invertible. – DanielWainfleet Jul 19 '21 at 07:01
3 Answers
0
You should specify the codomain of the function. After differentiating the function we understand that it is a strictly decreasing function.At x=1+ we get f(x) tends to infinity. x/(x-1)=1+1/x-1 ,so as x tends to infinity function tends to 1 Thus range is (1,inf).
Varad Tandale
- 19
- 2
0
If you compute $f'(x)=\frac{-1}{(x-1)^{2}}$ then you will see that it is $<0\,\forall\, x>1$ Hence you can say that it is one-one. And $f(x)=1+\frac{1}{x-1} > 1$ hence it is onto. So since it is a bijection on $(1,\infty)$ you can say that it is invertible.
Mr.Gandalf Sauron
- 13,906
0
Let x >1;
$f(x):=1+\dfrac{1}{x-1};$
$g(x):=x-1$ is stricly increasing.
$\dfrac{1}{g(x)}$ is strictly decreasing.
Peter Szilas
- 20,344
- 2
- 17
- 28