# Create a custom exponential function

I want to create a custom inverse exponential pdf with domain [0.00002, 0.0001] and range [0,1], where

x = 0.00002, y = 0 and

x = 0.0001, y = 1

The pdf should have the shape like that for a function like 1 - exp(-x).

How do I form this function?

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Y=F(x)? If so you can't have y of the form c(1-exp(-x))with y=0 at x=0.00002. –  Michael Chernick Aug 21 '12 at 23:43
What do you want to achieve? Can you be more specific as far as your data and requirements are concerned? –  Stat-R Aug 22 '12 at 0:46
@Stat-R: I don't want the function to be exactly like (1-exp(-x)). I just want the pdf to have the same curvature. –  Bruce Aug 22 '12 at 0:51
@Bruce. By curvature, do you mean shape? –  Stat-R Aug 22 '12 at 22:33
@Stat-R: Exactly –  Bruce Aug 22 '12 at 23:17

Adding to @MichaelChernick's comment, if y reaches 0 for a function like $y = c(1-e^{-tx})$ , $t$ must be 0 which reduces the function to $c$. The mapping you have desired for the above function is not correct logically unless you want y to be a constant function.

EDIT

If I understood you well...the family of curves that have shape like $log(1+x)$ and $(1-exp(-x))$ are just $a+b*log(1+c*x)$ and $a*(1-exp(-b*x))$ receptively.

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You are right. I did not realize that. But can't I create a function which has a curvature similar to ln(x) or (1-exp(-x)) and satisfies this property. Is there some family of functions which satisfy this property. –  Bruce Aug 22 '12 at 0:56