Computing the image of $(0, 1)$ under $f(x) = 1/(1 + x)$

Question

I would like to compute the image of $(0, 1)$ under $1/(1 + x)$.

I know that the answer is $(1/2, 1)$, but I would like to prove it. I am given that the function is strictly decreasing on the interval, and I am also provided that that the function is differentiable. So, I thought that this was hinting at taking a derivative, but I'm not sure.

Letting $f(x) = 1/(1 + x)$, I've computed $f'(x) = -1/(1 + x)^{2}$, which is always negative on $(0, 1)$, so it's decreasing on this interval. But, this doesn't tell me much since I already know the function is strictly decreasing.

Another approach I thought about taking was defining a set $S$ to be $(1/2, 1)$. Then, choosing an element $x\in \text{Im } f$ and showing that it's also in $S$ and vice-versa. I don't think this is the way I'm supposed to be doing it though, since the next part of the problem asks me to prove that the inverse of the function is $f^{-1}(y) = (1 - y)/y$.

Any help is appreciated.

Answer 1

The function is continuous on the interval and strictly decreasing thus since $f(0)=1$ and $f(1)=\frac12$ it takes all the values in between and we can conclude that the image is $(1/2,1)$.

Answer 2

A continuous function takes an interval to an interval.

Moreover, if the function is strictly monotonic, then it takes endpoints to endpoints.

The endpoints of our domain are $0$ and $1$.

Thus the endpoints of the image are $f(0)=1$ and $ f(1) =1/2$ and the image is $(1/2,1)$