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
you may see no benefit, but someone reading your posts would ... the whole idea is to effectively communicate mathematics.and for my questions I do not see the benefit in latex?
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.f:x -> 2x+1 g:x -> 1/x+1
1/(x+1)