# Math Help - Inverse function

1. ## Inverse function

I have been working on a project where I have a function F(x) which is closely related to a function G(x). The function G(x) has been well studied already and tools exist for using it. To make the equations I derive for F(x) useful I wanted to convert them into G(x) so that people can evaluate things for F(x) using the tools that already exist for G(x)
So far I derived their relationship
$r\cdot F(x)+ c= G(x)$

So for example if I had this equation to find an important number B
$B= F(5)-F(y)$

I could turn it into
$B= \frac{G(5)-c}{r}-\frac{G(y)-c}{r}$

$B= \frac{G(5)-G(y)}{r}$

And use the tools that exist for G(x) to find B for any y

But I am stuck when it comes to their inverse functions. As an example of this I have derived the equation
$L=F^{-1}(y-0.5)$

I am confused about how I can use the relationship $r\cdot F(x)+ c= G(x)$ to get L in terms of the inverse function of $G$. I am not even sure if this is possible to do

2. ## Re: Inverse function

Hey Shakarri.

One method I think you should investigate is the rule of the derivative of an inverse function (in terms of its non-inverse) and the chain rule of differentiation.

If you combine those together you should get a Taylor series for your function and if its in a simple enough form (or recognizable form), then you can substitute the appropriate elementary functions in.

3. ## Re: Inverse function

Hi !
if I well understand, you previously compute the values of c and r so that (G(x)-c)/r be a good approximation for F(x).
Then, you want to compute x so that F(x)=y , given y. (i.e. the inverse function of F)
G(x)=c+r*F(x) = c+r*y
Hence : x = value of the inverse function of G, for the argument X=(c+r*y)
That is : x = F^(-1){y} = G^(-1){X} where X=c+r*y

4. ## Re: Inverse function

Ah thank you, I was wondering about that solution but I was really unsure about inverse functions. Anyway now you have derived it and it works with all the cases which I can think of with trivial solutions so it must be correct