Find inverse composition

Question

I've been stuck for an hour trying to solve this composition function:

If $f(\sqrt[7] x)=g(3x-5)$, then $[(g^{-1}\circ f)^{-1} \circ g^{-1}](x)=\ldots$

My friend gave me a hint that I should find $f^{-1}$ first, but I don't really get it. Can someone help me to understand this problem?

I think the hint of your friend comes from: $(g^{-1}\circ f)^{-1}\circ g^{-1}=(f^{-1}\circ g)\circ g^{-1}=f^{-1}\circ(g\circ g^{-1})=f^{-1}$. But I don't see a way to find $f^{-1}$. — drhab, Jul 06 '17 at 15:51
im stuck at $f(\sqrt[7] {\frac {y+5}3}$ when i tried to let y=3x-5 but im not sure — Znik Dzulqarnain, Jul 06 '17 at 16:07
Is this a functional equation? It is not clear to me what is demanded. — Surb, Jul 06 '17 at 16:10
A hint could maybe be: $(f \circ g)^{-1} = g^{-1}\circ f^{-1}$ — mathreadler, Jul 06 '17 at 16:24
@Shuri2060: it is more complicated and probably harder to digest than what I wrote. — mathreadler, Jul 06 '17 at 16:26
More to read and make sense of increases the risk of mental block. — mathreadler, Jul 06 '17 at 16:28

Shuri2060 · Accepted Answer · 2017-07-06T17:06:15.023

We are given $f(x^{\frac{1}{7}})=g(3x-5)$ and asked to find ${(g^{-1}\circ f)}^{-1}\circ g^{-1}$.

Since you've been asked to find this, you can assume $g$ and $(g^{-1}\circ f)$ are invertible. Therefore $g \circ(g^{-1}\circ f)=f$ is invertible.

Now,

$$f(x^{\frac{1}{7}})=g(3x-5)\iff f(x)=g(3x^7-5)\iff f^{-1}(x)={\left(\frac{g^{-1}(x)+5}{3}\right)}^{1\over7}$$

These are equivalent statements, so you've essentially been given the last statement, and asked to find ${(g^{-1}\circ f)}^{-1}\circ g^{-1}=(f^{-1}\circ g)\circ g^{-1}=f^{-1}$.

Therefore you can choose any invertible $g(x)$, and this is as far as you can get.

Find inverse composition

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