1. ## inverse function proof

f:x -> 2x+1 g:x -> 1/x+1
show that {f[g(x)]}^-1 = g^-1[f^-1(x)]

i get {f[g(x)]}^-1 =x-3

and i get g^-1[f^-1(x)] = 4-2x//x-1

2. ## Re: inverse function proof

If $g(x)=\dfrac{1}{x+1}$ ...

$\{f[g(x)]\}^{-1}=g^{-1}[f^{-1}(x)]=\dfrac{3-x}{x-1}$

When are you going to start using LaTeX?

3. ## Re: inverse function proof

here is a picture of the question

Im stuck on question 1(iv) i understand how to get i,ii,iii but part iv will not work out for me
i get part iii to be x-3

and for my questions I do not see the benefit in latex?
If you really want me to use it though I will?

4. ## Re: inverse function proof

$f(g(x)) = 2\left(\dfrac{1}{x+1}\right)+1$

Solve for $x$:

$x = \dfrac{3-f(g(x))}{f(g(x))-1}$

So, $[f(g(x))]^{-1} = \dfrac{3-x}{x-1}$

So, you got part (iii) wrong. Skeeter understood and was telling you that you had it wrong.

5. ## Re: inverse function proof

its working out for me now. cheers

6. ## Re: inverse function proof

and for my questions I do not see the benefit in latex?
you may see no benefit, but someone reading your posts would ... the whole idea is to effectively communicate mathematics.

f:x -> 2x+1 g:x -> 1/x+1
if you're not going to use latex, at least use proper grouping symbols to make clear your expressions. you've been told this before.

1/(x+1)