Finding a smooth function on the half-open interval with certain properties I am looking to find a strictly increasing smooth function $r \colon [-1,0) \to \mathbb R$ with the following properties


*

*$r (-1) = c \in (1, \infty)$

*$r' (-1) = 0$

*as $t \to 0$, $r (t) \to \infty$ at the same rate as $1/t$.


I think for the first two conditions one could cook up a formula using the function $\exp (-1/t^2)$, but I am not sure how to satisfy the third condition.
(I am trying to plot a section of a line bundle over the circle that looks in one chart like the function $t \mapsto 1/t$ using the homeomorphism of the cylinder with the punctured plane — the function $r$ is supposed to play the role of the radius of the function.)
 A: Define $r$ on $[-1,0)$ as follows:
$$r(t)= c+e^{-1/(t+1)}-\frac{(1+t)^2}{t},\,\,-1<t<0,\,\,\,r(-1)=c. $$
Then everything is satisfied.
A: OK, so we need $r: [-1,0) \to \Bbb R$ such that:
$r(-1) = c \in (1, \infty), \tag 1$
$r'(-1) = 0, \tag 2$
$r(t) \to \infty \; \text{as} \; \dfrac{1}{t} \; \text{when} \; r \to 0^-; \tag 3$
It seems to me that one can start with a function which obviously satisfies (1) and (3) and then modify it, if necessary, so that (2) also holds; to this end, let $c \in (1, \infty)$ and 
$s(t) = -\dfrac{c}{t}; \tag 4$
then
$s(-1) = c, \tag 5$
and
$s'(t) = \dfrac{c}{t^2}, \tag 6$
so that
$s'(-1) = c; \tag 7$
we now modify $s(t)$ by adding a linear function $l(t)$ chosen so as not to disturb property (5), but adjusts (7) so that the slope of $s(t) + l(t)$ is $0$ at $t = -1$; this is possible since linear functions such as $l(t)$ are specified by their slope and value at one point, which we are free to choose; indeed, if we let 
$l(t) = c(t + 1) = ct + c, \tag 8$
then
$l(-1) = 0, \tag 9$
$l'(t) = c; \tag{10}$
we now set
$r(t) = s(t) - l(t) = -\dfrac{c}{t} - ct - c; \tag{11}$
then
$r(-1) = -\dfrac{c}{-1} -c(-1) - c = c + c - c = c, \tag{12}$
and
$r'(t) = s'(t) - c = \dfrac{c}{t^2} - c, \tag{13}$
whence
$r'(-1) = \dfrac{c}{(-1)^2} - c = 0; \tag{14}$
we note that
$r'(t) > 0, \; t \in (-1, 0); \tag{15}$
thus $r(t)$ is monotonically increasing in $(-1, 0)$,
$\displaystyle \lim_{t \to 0^-} r(t) = \infty, \tag{16}$
and
$\dfrac{r(t)}{1/t} = -c -ct^2 - ct \to -c \; \text{as} \; t \to 0^-, \tag{17}$
which I think shows that $r(t) \to \infty$ "at the same rate" as $1/t$, since they are asymptotically proportionate.
A: The function 
$$\frac{1}{t+2}-\frac{1}{t}$$
satisfies  the conditions 1. and 3., to also get 2. add a constant.
